LIBRARY 

OF  THE 

University  of  California. 


Class 


£-\.\'^  JKi\"  •        t\** 


X 


AN 


INTRODUCTION 


ALGEBRA 


UPON  THE 


INDUCTIVE  METHOD  OF  INSTRUCTION. 


BY  WARREN  COLBURN, 

AUTHOR  OF  INTELLECTUAL  ARITHMETIC  AND  SEQUEL  TO  DITTO. 


STEREOTYPED  AT  THE  BOSTON  TYPE  AND  STEREOTYPE  FOUHDRT, 
LATE  T.  H.  CARTER  AND  CO. 


CUMMIN GS,  HILLIARD,  AND  COMPANY. 
1826. 

Printed  at  TreadwelVs  Poicer  Press. 


c% 


DISTRICT  OF  MASSACHUSETTS,  to  wit. 

Dis^iet  Clerk's  Office. 
Be  it  remembered,  That  on  the  twenty-fourth  day  of  June,  A.  D.  1825,  in  the  for- 
ty-ninth year  of  the  Independence  of  the  United  States  of  America,  Warren  Colburn, 
of  the  said  district,  has  deposited  in  this  office  the  title  of  a  book,  the  right  whereof  he 
claims  as  author,  in  the  words  following,  to  wit : — 

'•  An  Introduction  to  Algebra,  upon  the  Inductive  Method  of  Instruction.  By  Warren 
Colburn,  Author  of  First  Lessons  in  Arithmetic,  &c/' 

In  conformity  to  the  act  of  the  Congress  of  the  United  States,  entitled  "  An  act  for  the 
encouragement  of  learning,  by  securuig  the  copies  of  maps,  charts,  and  tiooks,  lo  the 
authors  and  proprietors  of  such  copies,  during  the  times  therein  mentioned ;"  and  also  to 
an  act,  entitled  "  An  act  supplementary  to  an  act,  entitled  An  act  lor  the  encourage- 
ment of  learning,  by  securing  the  copies  of  maps,  charts,  and  books,  to  the  authors  and 
proprietors  of  such  copies,  during  the  times  therein  mentioned,  and  extending  the  bene- 
fits thereof  to  the  arts  of  designing,  engraving,  and  etching  historical  and  other  prints." 

JNO.  W.  DAVIS, 
Clerk  of  the  District  of  Massachusetts. 


PREFACE. 


Th£  first  object  of  the  author  of  the  foUowhig  treatise  has  been  to 
make  the  transition  from  arithmetic  to  algebra  as  gradual  as  possible. 
The  book,  therefore,  commences  with  practical  questions  in  simple 
equations,  such  as  the  learner  might  readily  solve  without  the  aid  of 
algebra.  This  requires  the  explanation  of  only  the  signs  plus  and 
minus,  the  mode  of  expressing  multiplication  and  division,  and  the 
sign  of  equality  ;  together  with  the  use  of  a  letter  to  express  the  un- 
known quantity.  These  may  be  understood  by  any  one  who  has  a 
tolerable  knowledge  of  arithmetic.  All  of  them,  except  the  use  of  the 
letter,  have  been  explained  in  arithmetic.  To  reduce  such  an  equation 
requires  only  the  application  of  the  ordinary  rules  of  arithmetic  ;  and 
these  are  applied  so  simply,  that  scarcely  any  one  can  mistake  them, 
if  left  entirely  to  himself.  One  or  two  questions  are  solved  first  with 
little  explanation  in  order  to  give  the  learner  an  idea  of  what  is  want- 
ed, and  he  is  then  left  to  solve  several  by  himself. 

The  most  simple  combinations  are  given  first,  then  those  which  are 
more  difficult.  The  learner  is  expected  to  derive  most  of  his  know- 
ledge by  solving  the  examples  himself ;  therefore  care  has  been  taken 
to  make  the  explanations  as  few  and  as  brief  as  is  consistent  with 
giving  an  idea  of  what  is  required. 

In  fact,  explanations  rather  embarrass  than  aid  the  learner,  because 
he  is  apt  to  trust  too  much  to  them,  and  neglect  to  employ  his  own 
powers  ;  and  because  the  explanation  is  frequently  not  made  in  the 
way,  that  would  naturally  suggest  itself  to  him,  if  he  were  left  to  ex- 
amine the  subject  by  himself.  The  best  mode,  therefore,  seems  to  be, 
to  give  examples  so  simple  as  to  require  little  or  no  explanation,  and 
let  the  learner  reason  for  himself,  taking  care  to  make  them  more  dif- 
ficult as  he  proceeds.  This  method,  besides  giving  the  learner  confi- 
dence, by  making  him  rely  on  his  own  powers,  is  much  more  interest- 
ing to  him,  because  he  seems  to  himself  to  be  constantly  making  new 
discoveries.  Indeed,  an  apt  scholar  will  frequently  make  original  ex- 
planations much  more  simple  than  would  have  been  given  by  the 
author. 

1.1.185-8 


4  Preface. 

This  mode  has  also  the  advantage  of  exercising  the  learner  in  rea- 
soning, instead  of  making  him  a  listener,  while  the  author  reasons  be- 
fore him. 

The  examples  in  the  first  fifty  pages  involve  nearly  all  the  opera- 
tions, that  are  ever  required  in  simple  numerical  equations,  with  one 
and  two  unknown  quantities. 

In  thQ  ninth  article,  the  learner  is  taught  to  generalize  particular 
cases,  and  to  form  rules.  Here  he  is  first  taught  to  represent  known 
quantities  by  letters,  and  at  the  same  time  the  purpose  of  it.  The 
transition  from  particular  cases  to  general  principles  is  made  as  gra- 
dual as  possible.  At  first  only  a  part  of  the  question  is  generalized, 
and  afterwards  the  whole  of  it. 

When  the  i^earner  understands  the  purpose  of  representing  known 
quantities  as  well  as  unknown,  by  letters  or  general  symbols,  he  is 
considered  as  fairly  introduced  to  the  subject  of  algebra,  and  ready 
to  commence  where  the  subject  is  usually  commenced  in  other  trea- 
tises. Accordingly  he  is  taught  the  fundamental  rules,  as  applied,  to 
literal  quantities.  Much  of  this  however  is  only  a  recapitulation  in 
a  general  form,  of  what  he  has  previously  learnt,  in  a  particular  form. 

After  this,  various  subjects  are  taken  up  and  discussed.  There  is 
nothing  peculiar  in  the  arrangement  or  in  the  manner  of  treating 
them.  The  author  has  used  his  own  language,  and  explained  as 
seemed  to  him  best,  without  reference  to  any  other  work.  A  large 
number  of  examples  introduce  and  illustrate  every  principle,  and  as 
far  as  seemed  practicable,  the  subjects  are  taught  by  example  rather 
than  by  explanation. 

The  demonstration  of  the  Binomial  Theorem  is  entirely  original,  so 
far  as  regards  the  rule  for  finding  the  coefiicients.  The  rule  itself  is 
the  same  that  has  always  been  used.  The  manner  of  treating  and 
demonstrating  the  principle  of  summing  series  by  difference,  is  also 
original.* 

Proportions  have  been  discarded  in  algebra  as  well  as  in  arithmetic. 
The  author  intended  to  give,  in  an  appendix,  some  directions  for  using 
proportions,  to  assist  those  who  might  have  occasion  to  read  other 
treatises  on  mathematics.  But  this  volume  was  already  too  large  to 
admit  it.  It  is  believed,  however,  that  few  will  find  any  difficulty  in 
this  respect.  If  they  do,  one  hour's  study  of  some  treatise  which  ex- 
plains proportions  will  remove  it. 

*  See  Boston  Journal  of  Philosophy  and  the  Arts,  No.  5,  for  May,  1835. 


Preface.  5 

In  order  to  study  this  work  to  advantage,  the  learner  should  solve 
every  question  in  course,  and  do  it  algebraically.  If  he  finds  a  ques- 
tion which  he  can  solve  as  easily  without  the  aid  of  algebra  as  with 
it,  he  may  be  assured,  this  is  what  the  author  expected.  If  he  first 
solves  a  question,  which  involves  no  difiiculty,  he  will  understand 
perfectly  what  he  is  about,  and  he  will  thereby  be  enabled  to  encoun- 
ter those  which  are  difficult. 

When  the  learner  is  directed  to  turn  back  and  do  in  a  new  way, 
something  he  has  done -before,  let  him  not  fail  to  do  it,  for  it  will  be 
necessary  to  his  future  progress  ;  and  it  will  be  much  better  to  trace 
the  new  principle  in  what  he  has  done  before,  than  to  have  a  new  ex- 
ample for  it. 

The  author  has  heard  it  objected  to  his  arithmetics  by  some,  that 
they  are  too  easy.  Perhaps  the  same  objection  will  be  made  to  this 
treatise  on  algebra.  But  in  both  cases,  if  they  are  too  easy,  it  is  the 
fault  of  the  subject,  and  not  of  the  book.  For  in  the  First  Lessons, 
there  is  no  explanation  ;  and  in  the  Sequel  there  is  probably  less  than 
in  any  other  books,  which  explain  at  all.  As  easy  however  as  they 
are,  the  author  believes  that  whoever  undertakes  to  teach  them,  will 
find  the  intellects  of  his  scholars  more  exercised  in  studying  them, 
than  in  studying  the  most  difficult  treatise  he  can  put  into  their 
hands.  When  the  learner  feels,  that  the  subject  is  above  his  capacity, 
he  dares  not  attempt  any  thing  himself,  but  trusts  implicitly  to  the 
author ;  but  when  he  finds  it  level  with  his  capacity,  he  readily  en- 
gages in  it.  But  here  there  is  something  more.  The  learner  is  re- 
quired to  perform  a  part  himself.  He  finds  a  regular  part  assigned 
to  him,  and  if  the  teacher  does  his  duty,  the  learner  must  give  a  great 
many  explanations  which  he  does  not  find  in  the  book. 
1* 


TABLE  OF  CONTENTS. 


Introduction.  Containing  a  brief  explanation  of  the  purpose  of 
algebra,  and  of  some  of  the  signs  9 

I.  Questions  producing  simple  equations,  in  which  the  unknown 
quantity  is  multiplied  by  known  quantities  11 

An  equation,  what  13 

Coefficients,  what  13 

II.  Questions  producing  simple  equations,  in  which  the  un- 
known quantity  is  divided  by  known  quantities  14 

III.  Questions  producing  simple  equations,  in  which  the  un- 
known quantity  is  added  to  known  quantities  19 

IV.  Questions  producing  simple  equations,  in  which  quantities 
consisting  of  two  or  more  terms  are  to  be  multiplied  by  a 
number  34 

Transposition  explained  26 

V.  Questions  producing  simple  equations,  in  which  quantities 
consisting  of  two  or  more  terms  are  to  hQ  divided  by  a  num- 
ber 30 

VI.  Questions  producing  simple  equations,  in  which  quantities 
consisting  of  two  or  more  terms,  some  of  them  having  the 
sign  —  before  them,  are  to  be  subtracted  from  other  quanti- 
ties 34 

Case  of  fractions  to  be  subtracted,  when  some  of  the  terms 
in  the  numerator  have  the  sign  —  36 

VII.  Examples  for  exercise  in  putting  questions  into  equation       37 
A  precept  useful  for  this  purpose  37 

VIII.  Questions  producing  equations  with  two  unknown  quan- 
tities 45 

Elimination,  what  47 

IX.  Explanation  of  some  of  the  higher  purposes  of  algebra,  and 
examples  of  generalization  51 

X.  Simple  quantities  66 

Addition,  multiplication,  and  subtraction  of  simple  quanti- 
ties 66 

XI.  Compound  Quantities  70 
Addition  of  compound  quantities                  ^  70 

XII.  Subtraction  of  compound  quantities  72 


Table  of  Contents.  7 

XIII.  Multiplication  of  compound  quantities  74 

XIV.  Division  of  algebraic  quantities  80 

XV.  Algebraic  fractions  82 
Multiplication  of  Algebraic  fractions  83 

XVI.  Division  of  algebraic  fractions  84 
To  multiply  fractions  by  fractions  66 

XVII.  Reducing  fractions  to  lower  terms  89 
Division  when  part  of  the  factors  of  the  divisw  are  con- 
tained in  the  dividend                                                            90 

XVIII.  Addition  and  subtraction  of  fractions  91 
To  reduce  fractions  to  a  common  denominator                       91 

XIX.  Division  of  whole  numbers  by  fractions  and  fractions  by 
fractions  '  95 

XX.  Division  of  compound  quantities  98 

XXI.  A  few  abstract  examples  in  equations  102 

XXII.  Miscellaneous  Questions  producing  simple  equations       104 

XXIII.  Questions  producing  simple  equations  involving  more 
than  two  unknown  quantities  107 

XXIV.  Negative  quantities,  explanation  of  them  1 12 

XXV.  Explanation  of  negative  exponents  121 

XXVI.  Examination  of  general  formulas,  to  see  what  values 
the  unknown  quantities  will  take  for  particular  suppositions 
made  upon  the  known  quantities  123 

XXVII.  Questions  producing  equations  of  the  second  degree     131 

XXVIII.  Extraction  of  the  second  root  133 

XXIX.  Extraction  of  the  second  root  of  fractions  142 

XXX.  Questions  producing  pure  equations  of  the  second  de- 
gree 145 

XXXI.  Questions  producing  pure  equations  of  the  third  de- 
gree 150 

Extraction  of  the  third  root  151 

XXXII.  Extraction  of  the  third  root  of  fractions  159 

XXXIII.  Questions  producing  pure  equations  of  the  third  de- 
gree 161 

XXXIV.  Questions  producing  affected  equations  of  the  second 
degree  163 

General  formula  for  equations  of  the  second  degree  174 

XXXV.  Demonstration  of  the  principle  that  every  equation  of 
the  second  degree  admits  of  two  values  for  the  unknown 
quantity  175 

Discussion  concerning  the  possible  and  impossible  values 
of  the  unknown  quantity,  also  of  the  positive  and  nega- 
tive values  of  it,  in  equations  of  the  second  degree  177 

XXXVI.  Of  powers  and  roots  in  general  182 

XXXVII.  Roots  of  compound  quantities  189 


8  Table  of  Contents, 

XXXVIII.  Extraction  of  the  roots  of  compound  quantities  of 
any  degree  193 

XXXIX.  Extraction  of  the  roots  of  numerical  quantities  of 
any  degree  195 

XL.  Fractional  exponents  and  irrational  quantities  197 

XLI.  Binomial  Theorem  202 

XLII.  Summation  of  series  by  differences  208 

XLIII.  Same  subject  continued  217 

XLIV.  Binomial  Theorem,  continued  from  Art.  XLI.  221 

XLV.  Continuation  of  the  same  subject  226 

XLVI.  Progression  by  difference,  or  Arithmetical  progression  228 

XLVII.  Progression  by  quotient,  or  Geometrical  progression  233 

XLVIII.  Logarithms  239 

XLIX.  Same  subject  continued  242 

L.  Logarithms  of  fractions  249 

LI.  Same  subject  continued  256 

LII.  Questions  relating  to  Compound  Interest  260 

LIII.  Same  subject  continued  264 

LIV.  Annuities  267 

Miscellaneous  Examples       '  268 


ALGEBRA. 


'!  Introduction. 

The  operations  explained  in  Arithmetic  are  sufficient  for 
the  solution  of  all  questions  in  numbers,  that  ever  occur ;  but 
it  is  to  be  observed,  that  in  every  question  there  are  two  dis- 
tinct things  to  be  attended  to  ;  first,  to  discover,  by  a  course 
of  reasoning,  what  operations  are  necessary  ;  and,  secondly,  to 
perform  those  operations.  The  first  of  these,  to  a  certain  ex- 
tent, is  more  easily  learnt  than  the  second ;  but,  after  the 
method  of  performing  the  operations  is  understood,  all  the  dif- 
ficulty in  solving  abstruse  and  complicated  questions  consists 
in  discovering  how  the  operations  are  to  be  applied. 

It  is  often  difficult,  and  sometimes  absolutely  impossible  to 
discover,  by  the  ordinary  modes  of  reasoning,  how  the  funda- 
mental operations  are  to  be  applied  to  the  solution  of  questions. 
It  is  our  purpose,  in  this  treatise,  to  show  how  this  difficulty 
may  be  obviated. 

It  has  been  shown  in  Arithmetic,  that  ordinary  calculations 
are  very  much  facilitated  by  a  set  of  arbitrary  signs,  called 
figures  ;  it  will  now  be  shown  that  the  reasoning,  previous  to 
calculation,  may  receive  as  great  assistance  from  another  set 
of  arbitrary  signs. 

Some  of  the  signs  have  already  been  explained  in  Arithme- 
tic ;  they  will  here  be  briefly  recapitulated. 

(==)  Two  horizontal  lines  are  used  to  express  the  words 
"  are  equal  to,^^  or  any  other  similar  expression. 

(+)  A  cross,  one  line  being  horizontal  and  the  other  per- 
pendicular, signifies  "  added  to.^^  It  may  be  read  and,  more, 
plus,  or  any  similar  expression ;  thus,  7  -[-  5  =  12,  is  read  7 
and  5  are  12,  or  5  added  to  7  i*  equxd  to  12,  or  7  plus  5  is  equal 
to  12.     Plus  is  a  Latin  word  signifying  more,  - 

( — )  A  horizontal  line,  signifies  subtracted  from.  It  is  some- 
times read  less  or  7ninus.   Minus  is  Latin,  signifying  less.    Thus 


10  Algebra.  I. 

14  —  6  =  8,  is  read  6  subtracted  from  14,  or  14  less  6,  or  14 
minus  6  is  equal  to  8. 

Observe  that  the  signs  +  and  —  affect  the  numbers  which 
they  stand  immediately  before,  and  no  others.  Thus 
14_-.6  +  8  =  16;  andl4  +  8  — 6  =  16; 
and  8  —  6  -f-  14  r=  16  ;  and,  in  fine,  —  6  +  8  -j-  14  =  16.  In 
all  these  cases  the  6  only  is  to  be  subtracted,  and  it  is  the 
same,  whether  it  be  first  subtracted  from  one  of  the  numbers, 
and  then  the  rest  be  added,  or  whether  all  the  others  be  added 
and  that  be  subtracted  at  last. 

(X)  (.)  An  inclined  cross,  or  a  point,  is  used  to  express 
multiplication  ;  thus,  5  X  3  =  15,  or  5. 3=1 5. 

(-^)  A  horizontal  line,  with  a  point  above  and  another  be- 
low it,  is  used  to  express  division.     Thus  15  -7-  3  =  6,  is  read 

15  divided  by  3  is  equal  to  5. 

But  division  is  more  frequently  expressed  in  the  form  of  a 
fraction  {Ariih,  Art.  XVI.  Part  II.),  the  divisor  being  made  the 
denominator,  and  the  dividend  the  numerator.  Thus  y  =  5, 
is  read  15  divided  by  3  is  equal  to  5,  or  one  third  of  15,  is  5, 
or  15  contains  3,  5  times. 

Example.     6  X  9  +  15  —  3  =  7  .  8  —  V  +  14. 
This  is  read,  9  times  6  and  1 5  less  3  are  equal  to  8  times  7  less 

16  divided  by  4,  and  14. 

To  find  the  value  of  each  side  ;  9  times  6  are  54  and  15  are 
69,  less  3  are  66.  Then  8  times  7  are  56,  less  16  divided  by 
4,  or  4  are  52,  and  14  more  are  66. 

In  questions  proposed  for  solution,  it  is  always  required  to 
find  one  or  more  quantities  which  are  unknown  ;  these,  when 
found,  are  the  answer  to  the  question.  It  will  be  found  ex- 
tremely useful  to  have  signs  to  express  these  unknown  quanti- 
ties, because  it  will  enable  us  to  keep  the  object  more  steadily 
and  distinctly  in  view.  We  shall  also  be  able  to  represent 
certain  operations  upon  them  by  the  aid  of  signs,  which  will 
greatly  assist  us  in  arriving  at  the  result. 

Algebraic  signs  are  in  fact  nothing  else  than  an  abridgment 
of  common  language,  by  which  a  long  process  of  reasoning  is 
presented  at  once  in  a  single  view. 

The  signs  generally  used  to  express  the  unknown  quantities 
above  mentioned  are  some  of  the  last  letters  of  the  alphabet,  as 
Xy  y^  z,  &c. 


I.  Equations.  11 

1.  1.  Two  men,  A  and  B,  trade  in  company,  and  gain  267 
dollars,  of  which  B  has  twice  as  much  as  A.  What  is  the  share 
of  each  ? 

In  this  example  the  unknown  quantities  are  the  particular 
shares  of  A  and  B. 

Let  X  represent  the  number  of  dollars  in  A's  share,  then  2  x 
will  represent  the  number  of  dollars  in  B's  share.  Now  these 
added  together  must  make  the  number  of  dollars  in  both  their 
shares,  that  is,  267  dollars. 

cc-f  2a?  =  267 
Putting  all  the  a?'s  together,  3  ^  =  267 

If  3  a:  are  267,  1  a:  is  i  of  267  in  the  same  manner  as  if  3 
oxen  were  worth  $267,  1  ox  would  be  worth  ^  of  it. 

x=^    69  =:  A's  share. 
2^7  =  178  =  B's  share. 

2.  Four  men.  A,  B,  C,  and  D,  found  a  purse  of  money  con- 
taining $325,  but  not  agreeing  about  the  division  of  it,  each 
took  as  much  as  he  could  get ;  A  got  a  certain  sum,  B  got  5 
times  as  much  ;  C,  7  times  as  much  ;  and  D,  as  much  as  B  and 
C  both.     How  many  dollars  did  each  get  ^ 

Let  X  represent  the  number  of  dollars  that  A  got ;  then  B 
got  5  X,  C  7a:,  and  D  (5  a?  -[-  7a;)  =  12  a?.  These,  added  toge- 
ther, must  make  $325,  the  whole  number  to  be  divided. 

x-{-6x-\-7x-^12x=:225 
Putting  all  the  a?'s  together,         25  a?  =  325 

x=:    13  =  A's  share. 
bx=    65=  B's     " 
'7x=    91=C's     " 
12a?=156i=:D's     " 

JSTote,  All  examples  of  this  kind  in  algebra  admit  of  proof. 
In  this  case  the  work  is  proved  by  adding  together  the  several 
shares.  If  they  are  equal  to  the  whole  sum,  325,  the  work  is 
right.  As  the  answers  are  not  given  in  this  work,  it  will  be 
well  for  the  learner  always  to  prove  his  results. 

In  the  same  manner  perform  the  following  examples. 

3.  Said  A  to  B,  my  horse  and  saddle  together  are  worth 
$130,  but  the  horse  is  worth  9  times  as  much  as  the  saddle. 
What  is  the  value  of  each  ? 

4.  Three  men.  A,  B,  and  C,  trade  in  company,  A  puts  in  a 
certain  sum,  B  puts  in  3  times  as  much,  and  C  puts  in  as  much 


12  Algebra,  I. 

as  A  and  B  both ;  they  gain  $656.     What  is  each  man's  share 
of  the  gain  ? 

5.  A  gentleman,  meeting  4  poor  persons,  distributed  60 
cents  among  them,  giving  the  second  twice,  the  third  three 
times,  and  the  fourth  four  times  as  much  as  the  first.  How 
many  cents  did  he  give  to  each  ? 

6.  A  gentleman  left  11000  crowns  to  be  divided  between 
his  widow,  two  sons,  and  three  daughters.  He  intended  that 
the  widow  should  receive  twice  the  share  of  a  son,  and  that 
each  son  should  receive  twice  the  share  of  a  daughter.  Re- 
quired the  share  of  each. 

Let  X  represent  the  share  of  a  daughter,  then  2x  will  repre- 
sent the  share  of  a  son,  &c. 

7.  Four  gentlemen  entered  into  a  speculation,  for  which 
they  subscribed  $4755,  of  which  B  paid  3  times  as  much  as  A, 
and  C  paid  as  much  as  A  and  B,  and  D  paid  as  much  as  B  and 
C.     What  did  each  pay  ? 

8.  A  man  bought  some  oxen,  some  cows,  and  some  sheep 
for  $1400  ;  there  were  an  equal  number  of  each  sort.  For 
the  oxen  he  gave  $42  apiece,  for  the  cows  $20,  and  for  the 
sheep  $8  apiece.     How  many  were  there  of  each  sort  ? 

In  this  exaiTiple  the  unknown  quantity  is  the  number  of  each 
sort,  but  the  number  of  each  sort  being  the  san.B,  one  charac- 
ter will  express  it. 

Let  X  denote  the  number  of  each  sort. 
Then  x  oxen,  at  $42  apiece,  will  come  to  42  a:  dolls.,  and  x 
cows,  at  $20  apiece,  will  come  to    20  x  dolls.,  and  x  sheep,  at 
$8  apiece,  will  come  to  8  a?  dolls.     These  added  together  must 
make  the  whole  price. 

42a?-|-20a;-f  8:rr=1400 
Putting  the  a:'s  together,     .     .     70  a:  =  1400 

Dividing  by  70, a:=:    20 

Ans,     20  of  each  sort. 

9.  A  man  sold  some  calves  and  sonle  sheep  for  $374,  the 
calves  at  $5,  and  the  sheep  at  $7  apiece ;  there  were  three 
times  as  many  calves  as  sheep.  How  many  were  there  of 
each  .'* 

Let  X  denote  the  number  of  sheep  ;  then  3  x  will  denote  the 
number  of  calves. 


I.  Equations,  1 3 

Then  x  sheep,  at  $7  apiece,  will  come  to  7  a?  dolls.,  and  3  x 
calves,  at  $5  apiece,  will  come  to  5  times  3  x  dolls.,  that  is, 
15  a:  dolls. 
These  added  together  must  make  the  whole  price. 
7a?+15a:rr374 
Putting  the  a?'s  together,  22  a?  =  374 
Dividing  by  22,  a?  r=    17  =  number  of  sheep. 

3a?=    51  =        "         calves. 

The  learner  must  have  remarked  by  this  time,  that  when  a 
question  is  proposed,  the  first  thing  to  be  done,  is  to  find,  by 
means  of  the  unknown  quantity,  an  expression  which  shall  be 
equal  to  a  given  quantity,  and  then  from  that,  by  arithmetical 
operations,  to  deduce  the  value  of  the  unknown  quantity. 

This  expression  of  equality  between  two  quantities,  is  called 
an  equation.  In  the  last  example,  7  a:  -|-  1 5  a:  =  374  is  an  equa- 
tion. 

The  quantity  or  quantities  on  the  left  of  the  sign  ==  are  called 
the  first  mejnber,  those  on  the  right,  the  second  member  of  the 
equation.  (7  a:-}-  15  a;)  is  the  first  member  of  the  above  equa- 
tion, and  374  is  the  second  member. 

Quantities  connected  by  the  signs  -f-  and  —  are  called  terms. 
7  X  and  15  a?  are  terms  in  the  above  equation. 

The  figure  written  before  a  letter  showing  how  many  times 
the  letter  is  to  be  taken,  is  called  the  coefficient  of  that  letter. 
In  the  quantities  7  a?,  15  a,  22  a? ;  7,  15, 22,  are  coefficients  of  a:. 

The  process  of  forming  an  equation  by  the  conditions  of  a 
question,  is  called  putting  the  question  into  an  equation. 

The  process  by  which  the  value  of  the  unknown  quantity  is 
found,  after  the  question  is  put  into  an  equation,  is  called  solv- 
ing or  reducing  the  equation. 

No  rules  can  be  given  for  putting  questions  into  equations ; 
this  must  be  learned  by  practice  ;  but  rules  may  be  found  for 
solving  most  of  the  equations  that  ever  occur. 

After  the  preceding  questions  were  put  into  equation,  the 
first  thing  was  to  reduce  all  the  terms  containing  the  unknown 
quantity  to  one  term,  which  was  done  by  adding  the  coeffi- 
cients. As  7  a?  -j-  15  a?  are  22  a?.  Then,  since  22  a?  =  374, 1  x 
must  be  equal  to  ^^  of  374.     That  is, 

TVhen  the  unknovm  quantity  in  one  member  is  reduced  to  one 
term,  and  stands  equal  to  a  known  quantity  in  the  otJier,  its  value  is 

2 


14  Algebra.  II. 

found  by  dividing  ike  knovm  quantity  by  the  coefficient  of  the  un- 
knoum  quantity, 

10.  A  man  bought  some  oranges,  some  lemons,  and  some 
pears,  for  1 56  cents  ;  the  oranges  at  6  cents  each,  the  lemons 
at  4  cents,  and  the  pears  at  3  cents  ;  there  was  an  equal  num- 
ber of  each  sort.     Required  the  number  of  each. 

11.  In. fencing  the  side  of  a  field,  the  length  of  which  was 
450  yards,  t^vo  workmen  were  employed  ;  one  fenced  9  yards, 
and  the  other  6  yards  per  day.  How  many  days  did  they 
work  .'' 

12.  Three  men  built  780  rods  of  fence  ;  the  first  built  9 
rods  per  day,  the  second  7,  and  the  third  5 ;  the  second  work- 
ed three  times  as  many  days  as  the  first,  and  the  third,  twice  as 
many  days  as  the  second.     How  many  days  did  each  work  ? 

13.  A  man  bought  some  oxen,  some  cows,  and  some  calves 
for  $348  ;  the  oxen  at  $38  each,  the  cows  at  $18,  and  the 
calves  at  $4.  There  were  three  times  as  many  cows  as  oxen, 
and  twice  as  many  calves  as  cows.  How  many  were  there  of 
each  sort  .'* 

14.  A  merchant  bought  a  quantity  of  flour  for  $132  ;  for  one 
half  of  it  he  gave  $5  per  barrel,  and  for  the  other  half  $7. 
How  many  barrels  were  there  in  the  whole  ? 

Let  X  denote  one  half  the  number  of  barrels. 

15.  From  t^^o  towns,  which  are  187  miles  apart,  two  travel- 
lers set  out  at  the  same  time  with  an  intention  of  meeting ;  one 
of  them  travels  at  the  rate  of  8,  the  other  of  9  miles  each  day. 
In  how  many  days  will  they  meet  .'* 

II.     1.  A  cask  of  wine  was  sold  for  $45,  which  was  only  | 

of  what  it  cost.     Required  the  cost. 

L^t  X  denote  the  cost. 

3x 
Three  fourths  of  x  may  be  written  Jo?  or  —     The  latter  is 

4 

preferable. 


Am,  $60. 


4 

:45 

4 

=  15 

x  = 

60 

n.  Equations.  16 

If  f  of  a?  comes  to  45,  then  _  must  come  to  J  of  45,  or 

4 
15,  and  x  will  be  4  times  15,  or  60. 
A  better  method. 

^  =  45  — Ji'/v     .---^ 

4 

3ic  =  45x4=180 
a?  =  60 

Observe,  that  —  is  the  same  as  i  of  3  x.     Now  if  a  of  3  «r 

is  45,  3^  itself  must  be  4  times  45,  or  180;  3 x  being  180,  x 
must  be  i  of  180,  which  is  60. 

2.  A  man,  being  asked  his  age,  answered,  that  if  its  half 
and  its  third  were  added  to  it  the  sum  would  be  88.  What 
was  his  age  ? 

Let  X  denote  his  age ;  then, 

a?4-^+^=83 
^2^3 

Reducing  the  terms  to  a  com-  )  6  x.S  x.2  a^_gg 
mon  denominator,  )  "6^     ~6       ^ 

Adding  them  together,  ^^^=88 

6 

}  of  11  X  being  88, 1 1  a:  will  be  6  times  88,  U  a:  =  528 
Dividing  by  11,  a?  =  48 

Am.  48  years. 

3.  If  f  of  a  hogshead  of  wine  cost  $65 ;  what  will  a  hogs- 
head cost  at  that  rate  ? 

4.  There  is  a  pole  i  and  i  under   water,  and  5  feet  out  of 

water  ;  what  is  the  length  of  the  pole?  « 

% 

Let  X  denot^  the  whole  length.  Then  ^-f-^  -f-  5  must  be 
equal  to  the  whole  length.     Hence, 


Reducing  to  a  common  denominator. 


2^3^ 


6x     Sx  .  2x  ,   c. 


46  Algebra,  TL. 

Adding  together,  y^^*^^ 

Since  the  two  members  are  equal,  if —  be  subtracted  from 

6 

both,  they  will  still  be  equal ;  hence, 

6 
and  0?  =  30  Ans,   30  feet. 

Proof,  One  half  of  30  is  15,  and  one  third  of  thirty  is  10. 
Now  30=  15  + 10  +  5. 

There  is  another  mode  of  reducing  the  above  equation  which 
in  most  cases  is  to  be  preferred.     It  is  the  same  in  principle. 

If  both  members  of  an  equation  be  multiplied  by  the  same 
number,  they  evidently  will  still  be  equal. 

In  the  equation, 

a;=f.+^+5. 

First  multiply  both  members  by  2,  the  denominator  of  one  of 
the  fractions,  and  it  becomes, 

207  =  0? +?5+ 10. 
^3^ 

Next  multiply  both  members  by  3,  the  denominator  of  the 

other  fraction,  and  it  becomes, 

6a:  =  3a?  +  2o?  +  30 

or  6  a?  =  5  0?  +  3^* 
Subtracting  5  x  from  both  members, 

^     a?  =  30  as  before. 

5.  In  an  orchard  of  fruit  trees  \  of  them  bear  apples,  a  of 
them  pears,  \  of  them  plums,  7  bear  peaches,  and  3  bear  cher- 
ries ;  these  are  all  the  trees  in  the  orchard.  How  many  are 
thera? 

6.  A  farmer,  being  asked  how  many  sheep  he  had,  answer- 
ed, he  had  them  in  four  pastures  ;  in  the  first  he  had  \  of  them, 
in  the  second  i,  in  the  third  |,  and  in  the  fourth  he  had  24 
sheep.     How  many  had  he  in  the  whole  ? 

7.  A  person  having  spent  |  and  i  of  his  money,  had  #26| 
left.     How  much  money  had  he  at  first  ? 

8.  A  man  driving  his  geese  to  market,  was  met  by  another, 
who  said  good  morrow,  master,  with  your  hundred  geese  ;  said 
he,  I  have  not  a  hundred,  but  if  I  had  as  many  more,  and  half 


II.  Equations.  17 

as  many  more,  and  two  geese  and  a  half,  I  should  have  a  hun- 
dred.    How  many  had  he  ? 

9.  A  and  B  having  found  a  bag  of  money,  disputed  about 
the  division  of  it.  A  said  that  ^  and  i  and  i  of  the  money 
made  $130,  and  if  B  could  tell  how  much  money  there  was,  he 
should  have  it  all,  otherwise  none  of  it.  How  much  money 
was  there  in  the  bag  ? 

10.  Upon  measuring  the  corn  produced  in  a  field,  being  96 
bushels,  it  appeared  that  it  had  yielded  only  one  third  part 
more  than  was  sown.     How  much  was  sown  f 

1 1 .  A  man  sold  96  loads  of  hay  to  two  persons  ;  to  the  first 
I,  and  to  the  second  |  of  what  his  stack  contained.  How  ma- 
ny loads  did  the  stack  contain  at  first  ? 

12^  A  and  B  talking  of  their  ages,  A  says  to  B  if  i,  i,  and  ^ 
of  my  age  be  added  to  my  age,  and  2  years  more,  the  sum  will 
be  twice  my  age.     What  was  his  age  ? 

13.  What  sum  of  money  is  that  whose  i,  i,  and  |  part  added 
together  amount  to  £9  ? 

14.  The  account  of  a  certain  school  is  as  follows  :  j\  of  the 
boys  learn  geometry,  f  learn  grammar,  j\  learn  arithmetic, 
■g\  learn  spelling,  and  9  learn  to  read.  What  is  the  number 
of  scholars  in  the  school  ? 

15.  There  is  a  fish  whose  head  weighs  9  lb.  his  tail  weighs 
as  much  as  his  head  and  half  his  body,  and  his  body  weighs  as 
much  as  his  head  and  tail  both.  What  is  the  weight  of  the 
fish  ? 

Represent  the  weight  of  the  body  by  oo. 

16.  There  is  a  fish  whose  head  is  4  inches  long,  the  tail  is 
twice  the  length  of  the  head,  added  to  f  of  the  length  of  the 
body,  and  the  body  is  as  long  as  the  head  and  tail  both.  What 
is  the  whole  length  of  the  fish  f 

17.  A  and  B  talking  of  their  ages,  A  says  to  B,  your  age  is 
twice  and  three  fifths  of  my  age,  and  the  sum  of  our  ages  is  54. 
What  is  the  age  of  each  ? 

18.  A  man  divided  $40  between  two  persons  ;  to  the  first  he 
gave  a  certain  sum,  and  to  the  second  only  |  as  much.  How 
much  did  he  give  to  each  ?  v 

Let  X  denote  the  share  of  the  first,  — will  denote  the  share 

5 

2* 


18  Algebm.  If. 

of  the  second.     These  added  together  must  make  $40. 

=  40 

5 

Multiplying  by  5,         5  a?  -|-  3  a:  =  200 

Adding  together,  8  :c  =  200 

Dividing  by  8,  a?  =  25  =  share  of  the  first. 

1?=  16  =         "         second. 
6 

19.  Three  persons  are  to  share  f  290  in  the  following  man- 
ner :  the  second  is  to  have  two  thirds,  and  the  third  three  fourths 
as  much  as  the  first.     What  is  the  share  of  each  f  ^ 

20.  A  farmer  wishes  to  mix  100  bushels  of  provender,-*  con- 
sisting of  rye,  barley,  and  oats,  so  that  it  may  contain  4  as 
much  barley  as  oats,  and  \  as  much  rye  as  barley.  How  much 
of  each  must  there  be  in  the  mixture  ^ 

21.  Divide  40  apples  between  two  boys  in  the  proportion  of 
3  to  2. 

The  proportion  3  to  2  signifies  that  the  second  will  have  | 
as  many  as  the  first. 

22.  A  gentleman  gave  to  3  persons  £98.  The  second  re- 
ceived five-eighths  of  tha  sum  given  to  the  first,  and  the  third 
one-fifth  of  what  the  second  had.     What  did  each  receive  ? 

23.  A  prize  of  $1280  wa?  divided  between  two  persons,  in 
the  proportion  of  9  to  7.     What  was  the  share  of  each  I 

24.  Three  men  trading  in  company,  put  in  money  in  the  fol- 
lowing proportion  ;  the  first  3  dollars  as  often  as  the  second  7, 
and  the  third  5.  They  gain  $960.  What  is  each  man's  share 
of  the  gain .'' 

Observe,  the  second  put  in  ^  of  what  the  first  put  in,  and  the 
third  put  in  f . 

25.  Three  men  traded  together  ;  the  first  put  in  $700,  the 
second  $450,  and  the  third  $950.  They  gained  $420.  What 
was  the  share  of  each  ? 

Observe,  the  second  put  in  4^^  =  4^=  tV  of  what  the  first 
put  in,  (fee. 


in.  Equations,  1-9 

III.  1.  Two  men,  A  and  B,  hired  a  pasture  together  for 
$55,  and  A  was  to  pay  $13  more  than  B.  What  did  each 
pay? 

Suppose  B  paid  x  dollars  ;  A  was  to  pay  13  dollars  more  ; 
therefore  he  paid  a?  +  13.  These  put  together  must  make  the 
whole  55  dollars. 

x-^x+ 1^  =  55 
Putting  the  x's  together, 

2x  +  lS=z55 
It  appears  that  2^  is  not  so  much  as  55  by  13,  therefore  tak- 
ing 13  from  55, 

2^  =  55  —  13 
2a?  =  42 
Dividing  Ijy  2,  a?  =  21  =  B's  share. 

B's  share  is  $21,  and  A's,  being  13  more,  is  $34, 
a: -h  13=  21  -I-  13  =  34  =  A's  share. 
Proof.     34  -j-  21  =:  55  the  whole  sum. 

2.  A  man  bought  a  horse  and  chaise  for  $300 ;  the  horse 
cost  $28  more  than  the  chaise.     What  was  the  price  of  each  ? 

3.  A  man  bequeathed  his  estate  of  $12000  to  his  son  and 
daughter ;  the  son  was  to  have  $2350  more  than  the  daughter. 
What  was  the  share  of  each  ? 

4.  A  father  who  has  three  sons,  leaves  them  16000  crowns. 
The  will  specifies  that  the  eldest  shall  have  2000  crowns  more 
than  the  second,  and  that  the  second  shall  have  1000  more 
than  the  youngest.     What  is  the  share  of  each  f 

Let  X  denote  the  number  of  crowns  in  the  share  of  the 
youngest,  then  x  -{■  1000  will  denote  the  share  of  the  second, 
and  a?  -j-  1000  +  2000  will  denote  the  share  of  the  eldest. 
These  added  together  must  make  the  whole  sum. 

'  x  +  x-{-  1000  +  x-{-  1000  -p  2000  =  16000 
Putting  together  the  x^s  and  the  numbers, 

3  a: +4000  =  16000 
It  appears  that  3  a?  is  not  so  much  as  16000  by  4000,  therefore 
subtracting  4000  from  16000, 

3a?  =  l«000  — 4000 
2x  =  12000 
Dividing  by  3,  a?  =   4000  =;=  share  of  the  youngest.. 

The  share  of  the  youngest  is  :K)0  crowns ;  add  to  this  1000, 
it  makes  5000,  the  share  of  the  second, 

X  +  1000  =  5000  =  share  of  the  second. 


20  Algebra.  IH. 

Add  2000  more,  it  makes  7000,  the  share  of  the  eldest, 
X  -h  1000  4-  2000  =  7000  =  share  of  the  eldest. 
"^     Proof,     The  several  shares  added  make  16000  crowns  which 
is  the  whole  estate. 

5.  A  draper  bought  three  pieces  of  cloth,  which  together  mea- 
sured 159  yards  ;  the  second  piece  was  15  yards  longer  than 
the  first,  and  the  third  was  24  yards  longer  than  the  second. 
What  was  the  length  of  each  ^ 

6.  A  gentleman  bequeathed  an  estate  of  $65000  to  his  wife, 
two  sons,  and  three  daughters.  The  wife  was  to  have  $2000 
less  than  the  elder  son,  and  $3000  more  than  the  younger  son; 
and  the  portion  of  each  of  the  daughters  was  $3500  less  than 
that  of  the  younger  son.     Required  the  share  of  each. 

The  1st  example  may  be  performed  differently.  Let  x  de- 
note the  number  of  dollars  paid  by  A  ;  B  paid  $13  less,  there- 
fore X —  13  will  represent  the  number  of  dollars  paid  by  B. 
These  added  together  must  make  the  whole. 

a: -j_  a:— 13=:  55 
Putting  the  a?'s  together,      2x  —  13=:55 
It  appears  that  2  a?  is  more  than  55  by  13,  therefore  add  13  to 
55  to  make  2  x, 

2  a;  =  55-1-  13 

2  ^  =  68 

Dividing  by  2,  a:  =  34  ==  A's  share. 

This  gives  A's  share  $34,  from  which  subtract  $13,  and  it  gives 

B's  share  $21,  as  before, 

x—l^  —  2\  :=B's  share. 

In  the  same  manner  perform  the  2d  and  3d.  The  4th  may 
be  solved  in  a  similar  manner. 

Let  the  elder  son's  share  be  represented  by  x.  The  second 
son's  share,  being  $2000  less,  will  be  x  —  2000.  The  young- 
er son's  share,  being  $1000  less  still,  will  hex  —  2000  —  1000. 
These  added  together  must  make  the  whole  sum. 

X  4-  a?— 2000  4-  a;— 2000—  1000  =  16000 
Putting  the  cc's  together  and  the  numbers  together, 

3  a:__  5000  =16000. 

It  appears  that  3  a:  is  more  than  16000  by  6000,  therefore  add 
5000  to  16000, 

3a?=  16000+ 5000 

3ic  =  21000 
Dividing  by  3,  x  =   7000 


III.  Equations.  21 

The  elder  son's  share  is  $7000,  as  before.  The  others  may 
be  easily  found  from  this. 

Again,  let  x  denote  the  second  son's  share.  The  elder  son's, 
being  $2000  more,  will  he  x  -\-  2000.  The  younger  son's, 
being  $1000  less,  will  be  x — 1000.  These  added  together 
must  make  the  whole. 

X  4-  2000  +  X  +  a:— 1000  =  16000 
Putting  the  oj's  together  and  the  numbers  together, 
3  a: +1000=  16000 

3a:  r=  16000— 1000 
337=15000 
X  =    5000 
The  second  son's  share  is  $5000,  as  before.     From  this  the 
rest  are  easily  found. 

Perform  the  5th  and  6th  in  a  similar  way. 

7.  At  a  certain  election  943  men  voted,  and  the  candidate 
chosen  had  a  majority  of  65.     How  many  voted  for  each  ? 

8.  A  person  employed  4  workmen  ;  to  the  first  of  whom  he 
gave  2  shillings  more  than  to  the  second  ;  to  the  second  3  shil- 
lings more  than  to  the  third  ;  and  to  the  third  4  more  than  to 
the  fourth.  Their  wages  amounted  to  32  shillings.  What  did 
each  receive  ? 

9.  A  cask,  which  held  146  gallons,  was  filled  with  a  mixture 
of  brandy,  wine,  and  water.  In  it  there  were  15  gallons  of 
wine  more  than  there  were  of  brandy,  and  as  much  water  as 
both  wine  and  brandy.     What  quantity  was  there  of  each  ? 

Observe,  that  after  the  question  is  put  into  equation,  the  pur- 
pose is  to  make  x  stand  alone  in  one  member  of  the  equation, 
equal  to  a  known  quantity  in  the  other  member,  then  the  value 
of  X  is  found.  In  the  preceding  examples  in  this  Art.  x  has 
been  found  only  in  the  first  member,  but  connected  with  known 
quantities  by  the  signs  -\-  and  — .  In  the  solution  of  these  equa- 
tions tJie  first  thing  was  to  unite  all  the  x's  into  one  term,  and  all  the 
known  quantities  into  another.  Then,  if  the  number  which  stood 
on  the  same  side  with  x,  had  the  sign  -(-  before  ity  that  number  was 
subtracted  from  the  other  member  of  the  equation  ;  but  if  it  had  the 
sign  —  before  it,  it  was  added  to  the  other  member.  Then  the  second 
member  was  divided  by  the  coefficient  of  x,  and  the  ansiver  was  ob- 
tained. 

10.  A  and  B  began  to  trade  with  equal  stocks.  In  the  first 
year  A  gained  a  sum  equ^l  to  twice  his  stock  and  £27  over  j 


22  Algebra.  III. 

B  gained  a  sum  equal  to  his  stock  and  £153  over.  Now  the 
amount  of  both  their  gains  was  equal  to  5  times  the  stock  of 
either.     What  was  the  stock  ? 

Let  X  denote  the  stock.  Then  A's  gain  was  2  a;  +  27,  and 
B's  was  a:  +  153.  These  added  together  must  make  5  times 
the  stock,  that  is,  5  x. 

5jc=2x  +  27-}-a;  + 153 
Uniting  the  x's  in  2d  member,  and  the  numbers, 

5  a?  =  3  a? +  180 
Subtracting  3  x  from  both  sides, 

2x=  180 
x=z    90 

11.  A  young  man  being  asked  his  age,  answered  that  if 
the  age  of  his  father,  which  was  44  years,  were  added  to  twice 
his  own,  the  sum  would  be  four  times  his  own  age.  What  was 
his  age  ^ 

12.  A  man  meeting  some  beggars,  gave  each  of  them  4 
pence,  and  had  16  pence  left ;  if  he  had  given  them  6  pence 
apiece,  he  would  have  wanted  12  pence  more  for  that  pur- 
pose. How  many  beggars  were  there,  and  how  much  mo- 
ney had  he  ^ 

Let  X.  represent  the  number  of  beggars. 

13.  A  man  has  six  sons,  each  of  whom  is  4  years  older  than 
his  next  younger  brother ;  and  the  eldest  is  three  times  as  old 
as  the  youngest.     Required  their  ages. 

14.  Three  persons,  A,  B,  and  C,  make  a  joint  contribution, 
which  in  the  whole  amounts  to  £76,  of  which  A  contri- 
butes a  certain  sum,  B  contributes  as  much  as  A  and  £10 
more,  and  C  as  much  as  A  and  Bboth.  Required  their  several 
contributions. 

15.  A  boy,  being  sent  to  market  to  buy  a  certain  quantity 
of  meat,  found  that  if  he  bought  beef,  which  was  4  pence  per 
pound,  he  would  lay  out  all  the  money  he  was  entrusted  with  ; 
but  if  he  bought  mutton,  which  was  3J  pence  per  pound, 
he  would  have  2  shillings  left.  How  much  meat  was  he  sent 
for  ? 

16.  A  man  lying  at  the  point  of  death  left  all  his  estate  to 
his  three  sons,  to  be  divided  as  follows  :  to  A  he  gave  one  half 
of  the  whole  wanting  $500  ;  to  B  one  third  ;  and  to  C  the  rest, 


III.    '^  Equations,  23 

which  was  f  100  less  than  the  share  of  B.    What  was  the  whole 
estate,  and  what  was  each  son's  share  ? 
Let  X  represent  the  whole  estate. 

A's  share  will  be  — —  500 

2 

B's  share     .     .    — 
3 

C's  share    .     .    — — 100 

These  together  will  be  equal  to  the  whole  estate,  which  was 
represented  by  x, 

-^  —  500 +^+-£.  —  100  =  a? 
2  3       3 

Uniting  j?'s  and  numbers  in  the  first  member, 

I:f_600=^ 
6  6 

—  is  greater  than  _?  by  600,  therefore 
6  6 

l5=^-h600 
6        6 

-^=600 
6 

X  =  3600 

The  whole  estate  is  $3600;  the  shares  are   f  1300,   $1200, 

and  $1100,  respectively. 

17.  A  father  intends  by  his  will,  that  his  three  sons  shall 
share  his  property  in  the  following  manner ;  the  eldest  is  to  re- 
ceive 1000  crowns  less  than  half  the  whole  fortune  ;  the 
second  is  to  receive  800  crowns  less  than  J  of  the  whole  ;  and 
the  third  is  to  receive  600  crowns  less  than  ^  of  the  whole. 
Required  the  amount  of  the  whole  fortune,  and  the  share  of 
each. 

18.  A  father  leaves  four  sons,  who  share  his  property  in  the 
following  manner  ;  the  first  takes  3000  livres  less  than  one  half 
the  fortune  ;  the  second,  1000  livres  less  than  one  third  of  the 
whole  ;  the  third,  exactly  one  fourth  ;  and  the  fourth  takes  600 
livres  more  than  one  fifth  of  the  whole.  What  was  the  whole 
fortune,  and  what  did  each  receive  ^ 


34  Algebra.  IV. 

19.  In  a  mixture  of  copper,  tin,  and  lead;  16  IK  less  than 
one  half  of  the  whole  was  copper  ;  12  lb.  less  than  one  third 
of  the  whole  was  tin,  and  4  lb.  more  than  one  fourth'  of  the 
whole  was  lead.  What  quantity  of  each  was  there  in  the  mix- 
ture ^ 

20.  A  general  having  lost  a  battle,  found  that  he  had  only 
3600  men  more  than  one  half  of  his  army  left,  fit  for  action ; 
600  more  than  one  eighth  of  them  being  wounded,  and  the  rest, 
which  amounted  to  one  fifth  of  the  whole  army,  either  slain  or 
taken  prisoners.  Of  how  many  men  did  his  army  consist  be- 
fore the  battle  ^ 

21.  Seven  eighths  of  a  certain  number  exceeds  four  fifths  of 
it  by  6.     What  is  that  number  i* 

22.  A  and  B  talking  of  their  ages,  A  says  to  B,  one  third  of 
my  age  exceeds  its  fourth  by  5  years.     What  was  his  age  f 

23.  A  sum  of  money  is  to  be  divided  between  two  persons, 
A  and  B,  so  that  as  often  as  A  takes  £9,  B  takes  £4.  Now  it 
happens  that  A  receives  £15  more  than  B.  What  is  the  share 
of  each  .f* 

24.  In  a  mixture  of  wine  and  cider,  25  gallons  more  than 
half  the  whole  was  wine,  and  5  gallons  less  than  one  third  of 
the  whole  was  cider.     How  many  gallons  were  there  of  each  ? 

IV.  1 .  A  man  having  some  calves  and  some  sheep,  and  be- 
ing asked  how  many  he  had  of  each  sort,  answered,  that  he  had 
20  more  sheep  than  calves,  and  that  three  times  the  number  of 
sheep  was  equal  to  seven  times  the  number  of  calves.  How 
many  were  there  of  each  ^ 

Let  X  denote  the  number  of  calves. 
Then  x-\-20  will  denote  the  number  of  sheep. 
7  times  the  number  of  calves  is  7  a?  ;  3  times  the  number  of 
sheep  is  3  jt?  -f  60  ;  for  it  is  evident  that  to  take  3  times  x  -|-  20, 
it  is  necessary  to  multiply  both  terms  by  3. 
By  the  conditions  these  must  be  equal, 
7a:  =  3a: +60. 
Subtracting  3  x  from  both  members, 
4a:  =  60 
a?  =:  15  =  number  of  calves, 
a?  -f-  20  =  35  =  number  of  sheep. 

Am,  15  calves,  and  35  sheep. 


IV.  Equations,  25 

2.  Two  men  talking  of  their  ages,  the  first  says,  your  age  is 
18  years  more  than  mine,  and  twice  your  age  is  equal  to  three 
times  mine.     Required  the  age  of  each. 

3.  Three  men,  A,  B,  and  C,  make  a  joint  contribution,  which 
in  the  whole  amounts  to  £276.  A  contributes  a  certain  sum, 
B  twice  as  much  as  A  and  £12  more,  and  C  three  times 
as  much  as  B  and  £12  more.  Required  their  several  con- 
tributions. 

4.  A  man  bought  7  oxen  and  11  cows  for  $591.  For  the 
oxen  he  gave  $15  apiece  more  than  for  the  cows.  How  much 
did  he  give  apiece  for  each  ? 

Let  X  denote  the  price  of  a  cow. 
Then  the  price  of  an  ox  will  be  a?  -|-  15. 
1 1  cows  at  X  dollars  apiece  will  come  to  1 1  a:  dollars. 
If  one  ox  cost  :c  -|-  1 5  dollars,  7  oxen  will  cost  7  times  x  -{- 
15,  which  is  7  jc  -|-  105. 

The  price  of  the  oxen  and  of  the  cows  added  together  will 
make  $591,  the  whole  price. 

nx  +  7x-{-  105  =  591 
Uniting  x%  1 8  a:  -|-  1 05  =  59 1 

Subtracting  105  from  both  members, 

18  a:  =  486 
Dividing  by  18,  cc  =  27  =  price  of  cows. 

a;  -(-  15  =  42  =  price  of  oxen. 

5.  A  man  bought  20  pears  and  7  oranges  for  95  cents.  For 
the  oranges  he  gave  2  cents  apiece  more  than  for  the  pears. 
What  did  he  give  apiece  for  each  ? 

6.  A  man  bought  20  oranges  and  25  lemons  for  $1.95.  For 
the  oranges  he  gave  3  cents  apiece  more  than  for  the  lemons. 
What  did  he  give  apiece  for  each  ? 

7.  Two  persons  engage  at  play,  A  has  76  guineas,  and  B  52, 
before  they  begin.  After  a  certain  number  of  games  lost  and 
won  between  them,  A  rises  with  three  times  as  many  guineas 
as  B.     How  many  guineas  did  A  win  of  B  ? 

Let  X  denote  the  number  of  guineas  that  A  won  of  B. 

Then  A,  having  gained  x  guineas,  will  have  76  -j-  a? 

B,  having  lost  x  guineas,  will  have  only         52  —  x 

A  has  now  three  times  as  many  as  B,  that  is,  3  times  52  —  x, 
which  is  156 —  3  a:.     It  is  evident  that  both  52  and  x  must  be 
multiplied  by  3,  because  52  is  a  number  too  large  by  x,  there- 
fore 3  times  52  will  be  too  large  by  3  x, 
3 


26  Algebra.  IV. 

76-f.a;=  156— 3a? 

a?=  156  — Sa?  — 76 
a?  4-3  a?  =  156—76 
4  a?  =  156— 76 
•      4x  =    80 
.     a:  ==:     20 

^n5.  20  guineas. 

Proof.  If  A  won  20  guineas  of  B,  A  will  have  96  and  B  32. 
3  times  32  are  96. 

This  equation  is  rather  more  difficult  to  solve  than  any  of  the 
preceding.  In  the  first  place  I  subtract  76  from  both  mem- 
bers, so  as  to  remove  it  from  the  first  member.  Then  to  get 
3  X  out  of  the  second  member,  which  is  there  subtracted,  I  add 
3  X  to  both  members  ;  then  the  a?'s  are  all  in  the  first  member, 
and  the  known  numbers  in  the  other. 

N.  B.  Any  term  which  has  the  sign  -\-,  either  expressed  or 
understood,  may  be  removed  from  one  member  to  the  other  by 
giving  it  the  sign  — ;  for  this  is  the  same  as  subtracting  it  fl-om 
both  sides.  Thus  a:-[-3=:10;  a?  is  not  so  much  as  10  by  3, 
we  therefore  say  a?  =  10  —  3.  Again,  5a?=^  18-f-3a?.  Now 
5  X  is  more  than  IS  by  3a?,  therefore  we  may  say  5  a?  —  3  a?  = 
18. 

Any  term  which  has  the  sign  —  before  it  may  be  removed 
from  one  member  to  the  other  by  giving  it  the  sign  +.  This 
is  equivalent  to  addii;ig  the  number  to  both  sides.  Thus  5  a? 
—  3  =  17.  In  this  it  appears  that  5  a;  is  more  than  17  by  3  ; 
therefore  we  say  5  a?  =  17  -|-  3.  Again,  5  a?  =  32  —  3  a?.  Here 
it  appears  that  5  a?  is  not  so  much  as  32  by  3a? ;  therefore  we 
say  5  a?  -j-  3  a;  =  32.     This  is  called  transposition. 

Hence  it  appears  that  any  term  may  be  transposed  from  one  mem- 
her  to  the  other,  care  being  taken  to  change  the  sign. 

In  the  last  example,  76  was  transposed  from  the  first  member 
to  the  second,  and  the  sign  changed  from  -f-  to  — ;  and  3  a? 
was  transposed  from  the  second  member  to  the  first,  and  the 
sign  changed  from -^  to  +.  This  has  been  done  in  many  of 
the  preceding  examples. 

When  a  number,  consisting  of  two  or  more  terms,  is  to  be  multi- 
plied, all  the  terms  must  be  multiplied,  and  their  signs  preserved.  In 
the  last  example,  52  —  x,  multiplied  by  3,  gave  a  product  156  — 
3a?. 

8.  A  person  bought  two  casks  of  wine,  one  of  which  held 
exactly  three  times  as  much  as  the  other.     From  each  he  drew 


IV.  Equations.  27 

4  gallons,  and  then  there  were  four  times  as  many  gallons  re- 
maining in  the  larger  as  in  the  smaller.  How  many  gallons 
were  there  in  each  at  first  ? 

Let  X  denote  the  number  of  gallons  in  the  less  at  first. 
Then  the  number  of  gallons  in  the  greater  will  be  3  x. 
Taking  4  gallons  from  each,  the  less  will  he  x  —  4 
And  the  greater         ....         3  a? — 4 
The  greater  is  now  4  times  as  large  as  the  less  5  4  times  x 
—  4  is  4^7 —  16. 

4a:— 16  =  30? —   4 
By  transposing  16,  4a?  =  3a?+16  —  4 

By  transposing  Sx,      4x — 3x=:z  16  —  4 
Uniting  terms,  a;  =  12  =  less. 

3  a?  rr:  36  =  greater. 
».^iis.  Less  12  gallons,  greater  36  gallons. 
Proof.     36  is  three  times  12  according  to  the  conditions. 
Take  4  from  each,  then  one  contains  32  and  the  other  8.     32  is 
4  times  8. 

9.  A  man  when  he  was  married  was  three  times  as  old  as  his 
wife  ;  after  they  had  lived  together  15  years,  he  was  but  twice 
as  old.     How  old  was  each  when  they  were  married  ? 

10.  A  farmer  has  two  flocks  of  sheep,  each  containing  the 
same  number.  From  one  of  these  he  sells  39,  and  from  the 
other  93  ;  and  finds  just  twice  as  many  remaining  in  the  one  as 
in  the  other.     How  many  did  each  flock  originally  contain  ? 

1 1 .  A  courier,  who  travels  60  mile^  per  day,  had  been  de- 
spatched 5  days,  when  a  second  was  sent  to  overtake  him ;  in 
order  to  which,  he  must  go  75  miles  per  day  ;  in  what  time  will 
he  oveirtake  the  former  ? 

12.  A  and  B  engaged  in  trade,  A  with  £240,  and  B  with 
£96.  A  lost  twice  as  much  as  B  ;  and  upon  settling  their  ac- 
counts it  appeared  that  A  had  three  times  as  much  remaining 
as  B.     How  much  did  each  lose  ^ 

Let  X  denote  B's  loss,  then  96  —  x  will  denote  what  he  had 
remaining.  2  a;  will  denote  A's  loss,  and  240  —  2x  what  he 
had  remaining,  &,c. 

13.  Two  persons  began  to  play  with  equal  sums  of  money ; 
the  first  lost  14  shillings,  and  the  other  won  14  shillings,  and 
then  the  second  had  twice  as  many  shillings  as  the  first.  What 
sum  had  each  at  first  ^ 


28  Algebra.  IV. 

14.  Says  A  to  B,  I  have  5  times  as  much  money  as  you;  yes, 
says  B,  but  if  you  will  give  me  $17,  I  shall  have  7  fimes  as 
much  as  you.     How  much  had  each  ? 

15.  Two  men,  A  and  B,  commenced  trade  ;  A  had  $500 
less  than  3  times  as  much  money  as  B ;  A  lost  $1500,  and  B 
gained  $900,  then  B  had  twice  as  much  as  A.  How  much  had 
each  at  first  ? 

16.  From  each  of  15  coins  an  artist  filed  the  value  of  2  shil- 
lings, and  then  offered  them  in  payment  for  their  original  value ; 
but  being  djetected,  the  whole  were  found  to  be  worth  no  more 
than  $145.     What  was  their  original  value  ? 

17.  A  boy  had  41  apples,  which  he  wished  to  divide  between 
three  companions,  as  follows  ;  to  the  second  he  wished  to  give 
twice  as  many  as  to  the  first,  and  three  apples  more ;  and  to 
the  third  he  wished  to  give  three  times  as  many  as  to  the 
second,  and  2  apples  more.  How  many  must  he  give  to 
each  ? 

18.  A  person  buys  12  pieces  of  cloth  for  149  crowns  :  2  are 
white,  3  are  black,  and  7  are  blue.  A  piece  of  the  black  costs 
2  crowns  more  than  a  piece  of  the  white,  and  a  piece  of  the 
blue  costs  3  crowns  more  than  a  piece  of  the  black.  Required 
the  price  of  each  kind. 

See  example  4th  of  this  Art. 

19.  A  man  bought  6  barrels  of  flour  and  4  firkins  of  butter  ; 
he  gave  $2  more  for  a  firkin  of  butter,  than  for  a  barrel  of  flour  ; 
and  the  butter  and  flour  both  cost  the  same  sum.  What  did 
he  give  for  each  f 

20.  A  grocer  sold  his  brandy  for  25  cents  a  gallon  more  than 
his  wine,  and  37  gallons  of  his  wine  came  to  as  much  as  32 
gallons  of  his  brandy.     What  was  each  per  gallon  f 

21.  A  man  bought  7  oxen  and  36  cows  ;  he  gave  $18  apiece 
more  for  the  oxen  than  for  the  cows,  and  the  cows  came  to 
three  times  as  much  as  the  oxen  wanting  $3.  What  was  the 
price  of  each  ? 

22.  A  man  sold  20  oranges,  some  at  4  cents  apiece,  and  some 
at  5  cents  apiece,  and  the  whole  amounted  to  90  cents.  How 
many  were  there  of  each  sort  ? 

If  he  had  sold  13  at  5  ceAts  apiece,  then  the  number  sold  at 
4  cents  apiece  would  be  20 — 13,  or  7. 


IV.  Equations.  29 

In  the  same  manner,  if  he  sold  x  oranges  at  5  cents  apiece, 
then  he  sold  20  —  x  oranges  at  4  cents  apiece,     x  oranges  at 
5  cents  apiece  would  come  to  5  a?  cents,  and  20  —  x  oranges  at 
4  cents  apiece  would  come  to  4  times  20  —  .r  cents,  which  is 
80  -«-  4  a?  cents. 
These  added  t02;ether  must  make  90  cents,  therefore 
5  a? .+  80— 4^  =  90 
By  transposing  80  and  uniting  terms,  a:  =  10  at  5  cents. 

Ans.  10  of  each  sort. 

23.  A  man  dying  left  an  estate  of  $2500  to  be  divided  be- 
tween his  two  sons,  in  such  a  manner,  that  twice  the  elder  son's 
share  should  be  equal  to  three  times  the  share  of  the  second. 
Required  the  share  of  each. 

Let  X  denote  the  younger  son's  share. 
Then  2500  —  x  will  denote  the  elder  son's  share. 
Twice  the  elder  son's  share  is  5000  —  2x. 
By  the  conditions,      S  x  =:  5000  —  2x 
By  transposition,         5  x  =:  5000 
Dividing  by  5,  x  =  1000 

2500 -—1000  =  1500 

Ans.  Elder  son  $1500,  younger  son  $1000. 

24.  Two  robbers,  after  plundering  a  house,  found  they  had 
35  guineas  between  them  ;  and  that  if  one  of  them  had  4  gui- 
neas more,  he  would  have  twice  as  many  as  the  other.  How- 
many  had  each  ? 

25.  A  man  sold  45  barrels  of  flour  for  $279  ;  some  at  $5 
and  some  at  $8  per  barrel.  How  many  barrels  were  there 
of  each  sort  ? 

26.  A  man  sold  some  oxen  and  some  cows  for  $330 ;  the 
whole    number  was  15.     He  sold  the  cows  for  $17  apiece,  '^ 
and  the  oxen  for  $32  apiece.     How  many   were   there   of 
each  sort  ? 

27.  After  A  had  lost  10  guineas  to  B,  he  wanted  only  8 
guineas  in  order  to  have  as  much  money  as  B  ;  and  together 
they  had  60  guineas.     What  money  had  each  at  first  ? 

Let  X  he  the  nuihber  of  guineas  A  had. 
Then  60 — a?  will  be  the  number  B  had. 
A  lost  10  to  B,  therefore  A's  is  diminished  by  10,  and  B's 
increased  by  10,  which  makes  A's  x  —  10,  and  B's  70  —  x. 
3* 


t 


30  Algebra.  V. 

By  the  conditions,    x  —  104-8  =  70  —  x 
Transposing  and  uniting,       2  a:  =  72 

a:  =  36  =  what  A  had. 
60  — 36  =  24  =  what  B  had. 

28.  Divide  the  number  197  into  two  such  parts, thaf  four 
times  the  greater  may  exceed  five  times  the  less  by  50. 

29,  Two  workmen  were  employed  together  for  50  days,  at 
5  shillings  per  day  each.  A  spent  6  pence  a  day  less  than  B 
ditr,  and  at  the  end  of  the  50  days  he  found  he  had  saved  twice 
as  much  as  B,  and  the  expense  for  two  days  over.  What  did 
each  spend  per  day  ^ 

Let  X  denote  what  A  spent  per  day  (in  pence). 

Then  60  —  x  (5s.  being  60d.)  will  be  what  he  saved  per 
day. 

B  saved  6d.  less  than  A. 

Therefore  54  —  x  will  be  what  B  saved  per  day. 

Multiplying  both  by  50,  the  number  of  days, 

A  saved  3000  —  50  x,  and  B  saved  2700  —  50  x. 

By  the  conditions  A  saved  2  x  more  than  twice  what  B 
saved. 

Therefore  3000  —  50  a:  =  5400  —  100  a:  +  2  a?    , 

Transposing  and  uniting,  48  a?  =  2400 

a?  z=    50    =  what  A  spent. 
50  -f  6  =:    56    =  what  B  spent. 

V.  1.  Two  persons  talking  of  their  ages,  A  said  he  was 
25  years  older  than  B,  and  that  one  half  of  his  age  was  equal 
to  three  times  that  of  B  wanting  35  years.  What  was  the  age 
of  each  ^ 

Let  X  denote  the  age  of  B. 

Then  the  age  of  A  will  be  a?  +  25. 

J  of  a:  +  25  is  expressed     "Li — 


Hence  we  have 


3a:— 35=  ^  +  ^^ 

2 

Multiplying  by  2,  6  a:  —  70  =  a:  +  25 

By  transposing  x  and  —70,  ^  ^  —  a?  =  25+  70 

Uniting  terms,  5  a?  =  95 

Dividing  by  5,  a:  =  19  =  B's  age. 

a:  +  25  =  44  =  A's  age. 


V.  Equations*  31 

JVote,  Since  J  of  a?  +  25  is  3  a:  —  35,  a?  +  25  must  be  twice 
3^ — 35. 

2.  Two  men  talking  of  their  horses,  A  says  to  B,  my  horse 
is  worth  $25  more  than  yours,  ai^d  |  of  the  value  of  my 
horse  is  equal  to  f  of  the  value  of  yours.  What  is  the  value 
of  each  f 

Let  X  denote  the  value  of  B's  horse. 
Then  the  value  of  A's  will  be  a?  -|-25. 

I  of  a;  +  25  is  ^  +  ^^,  f  is  3  times  as  much,  that  is  if-il^: 
5  5 

By  the  conditions,  Sx_Sx  +  75 

4  5 

Mtiltiplying  by  5,  1^  =  3  a?  -|-  75 

Multiplying  by  4,  15  a?  =  12  a?  +  300 

3a:  =300 
x=  100 

Ans.  A's  $125,  B's  $100. 
Proof.    The  first  condition  is   evidently  answered.     With 
regard  to  the  second,  |  of  125  is  75,  and  |  of  100  is  75. 

3.  Two  men  talking  of  their  ages,  one  says,  my  age  is  now 
I  of  yours,  but  in  twenty  years  from  this  time,  if  we  live,  it  will 
be  f  of  yours.     Required  the  age  of  each. 

Suppose  the  age  of  the  elder  x. 

Sx 
Then  the  younger  will  be  — 

In  20  years  the  age  of  the  elder  will  be  a?  -f  20,  and  of  the 
younger  ?^+ 20. 

By  the  conditions       ^^  +  ^^  =  ^  -|-  20 

5  4 

Multiplying  by  5,        4 a?  +  80  =  —-+  100 

4 

Multiplying  by  4,    16  a?  -f  320  =  15  a?  -f  400 

'^'320?''"'  ]  16^- 15^  =  400-320 

X  =•   80  =  age  of  elder. 

Sx 

—  =:   60  =  age  of  younger. 


S2  ^igebta,  ¥, 

4.  A  man  being  asked  the  value  of  his  horse  and  chaise, 
answered,  that  the  chaise  was  worth  $50  more  than  the 
horse,  and  that  one  half  of  the  value  of  the  horse  was  equal  to 
one  third  of  the  value  of  the  chaise.  Required  the  value  of 
each. 

5.  Two  persons  talking  of  their  ages,  the  first  says,  |  of  my 
age  is  equal  to  f  of  yours  ;  and  the  difference  of  our  ages  is  10 
years.     What  are  their  ages  ? 

6.  There  ar6  two  towns  situated  at  unequal  distances  from 
Boston,  and  on  the  same  road.  They  are  30  miles  apart.  | 
of  the  distance  of  the  second  from  Boston  is  equal  to  f  of 
the  distance  of  the  first.  What  is  the  distance  of  each  from 
Boston  ? 

7.  A  man  being  asked  the  value  of  his  horse  and  saddle,  an- 
swered, that  his  horse  was  worth  $114  more  than  his  saddle, 
and  that  |  of  the  value  of  his  horse  was  7  times  the  value  of 
his  saddle.     What  was  the  value  of  each  ? 

8.  A  hare  is  40  rods  before  a  greyhound,  but  she  can  run 
only  I  as  fast  as  the  greyhound.  How  far  will  each  of  them 
run  before  the  greyhound  will  overtake  the  hare  ? 

9.  A  gentleman  paid  4  labourers  $136  ;  to  the  first  he  paid 
3  times  as  much  as  to  the  second  wanting  $4 ;  to  the  third  one 
half  as  much  as  the  first,  and  $6  more  ;  and  to  the  fourth  4 
times  as  much  as  to  the  third,  and  $5  more.  How  much  did  he 
pay  to  each  ? 

10.  A  man  bought  some  cider  at  $4  per  barrel,  and  some 
beer  at  $7.  There  were  6  barrels  more  of  the  cider  than  of 
the  beer ;  and  |  of  the  price  of  the  beer  was  equal  to  J  of  the 
price  of  the  cider.     Required  the  number  of  barrels  of  each. 

1 1 .  Two  men  commenced  trade  together  ;  the  first  put  in 
£40  more  than  the  second^  and  the  stock  of  the  first  was  to  that 
of  the  second  as  14  to  5.     What  was  the  stock  of  each  ? 

14  to  5  signifies  the  second  is  -^^  of  the  first. 

12.  A  man's  age  when  he  was  married  was  to  that  of  his 
wife  as  3  to  2  ;  and  when  they  had  lived  together  4  years,  his 
age  was  to  hers  as  7  to  5.  What  were  their  ages  whfeh  they 
were  married  ? 

13.  A  and  iB  began  trade  with  equal  sums  of  money.  In  the 
first  year  A  gained  £40,  and  B  lost  £40  ;  but  in  the  second, 
A  lost  one  third  of  what  he  then  had,  and  B  gained  a  sum  less 


V.  Equatiom.  33 

by  £40  than  twice  the  sum  A  had  lost ;  when  it  appeared  that 
B  had  twice  as  much  money  as  A.  What  money  did  each  be- 
gin with  ? 

Let  X  be  the  number  of  pounds  each  had  at  first.  Then  x 
-f-  40  will  be  the  sum  A  had  at  the  end  of  the  first  year  ;  and 
X  —  40  the  sum  B  had. 

The  second  year  A  lost  J  of  what  he  then  had,  consequently 

2  a:  -4-  80 
he  saved  | ;  his  sum  will  then  be  ^I- 

o 

B  gained  twice  as  much  as  A  lost  wanting  £40 ;  his  will  be 

X  — 40  + ±. —  40. 

B  had  now  twice  as  much  as  A, 

4a^+160_^_^         2.r  +  80_^^ 
3  ^         3 

Multiplying  by  3, 

4a?  -I-  160  =  3a:—  120  -f  2  a:  -f  80  —  120. 
Transposing  and  uniting, 

-^x=  —  320. 
Transposing  again,  320  =z  a?, 

Ans,  £320. 
'  JYote.  In  this  example  the  result  had  the  sign  —  in  botfi 
members,  but  by  transposing  it  has  the  sign  -f-.  It  would 
have  been  the  same  thing  if  the  signs  had  been  changed  with- 
out transposing.  The  result  would  have  come  out  right  if  the 
first  member  had  been  made  the  second,  and  the  second  first, 
in  the  first  equation. 

14.  A  person  playing  at  cards,  cut  the  pack  in  such  a  man- 
ner, that  I  of  what  he  cut  off  were  equal  to  f  of  the  remainder. 
How  many  did  he  cut  off  .'^ 

15.  Divide  $183  between  two  men,  so  that  4  of  what  the  first 
receives,  shall  be  equal  to  j\  of  what  the  second  receives. 
What  will  be  the  share  of  each  ? 

16.  A  man  sold  20  bushels  of  grain,  rye  and  wheat ;  the  rye 
at  5s.  and  the  wheat  at  7s.  per  bushel  ;  |  of  the  rye  came  to 
as  much  as  4  of  the  wheat.     How  much  was  there  of  each  ? 

17.  What  number  is  that  from  which  if  5  be  subtracted  two 
thirds  of  the  remainder  will  be  40  ? 

18.  A  man  has  a  lease  for  99  years  ;  and  being  asked  how 


34  Mgcbra,  -     Vt. 

much  of  it  was  already  expired,  answered,  that  two  thirds  of 
the  time  past  was  equal  to  four  fifths  of  the  time  to  come.  Re- 
quired the  time  past,  and  the  time  to  come. 

19.  It  is  required  to  divide  the  number  50  into  two  such 
parts,  that  three  fourths  of  one  part  added  to  fi:ve  sixths  of  the 
other  may  make  40. 

20.  Two  workmen  received  equal  sums  for  their  work  ;  but 
if  one  of  them  had  received  18  dollars  more,  and  the  other  3 
dollars  less,  then  |  of  the  wages  of  the  latter  would  have  been 
equal  to  J  of  the  wages  of  the  former.  How  much  did  each 
receive  ? 

21.  A  certain  man,  when  he  married,  found  that  his  age  was 
to  that  of  his  wife  as  7  to  5 ;  if  they  had  been  married  8  years 
sooner,  his  age  would  have  been  to  hers  as  3  to  2.  What  were 
their  ages  at  the  time  of  their  marriage  ? 

VI.  1.  Divide  the  number  68  into  two  such  parts,  that  the 
difference  between  the  greater  and  84,  may  be  equal  to  three 
times  the  excess  of  40  above  the  less. 

Let  X  =1  the  less. 

Then  68  —  x=.  the  greater. 

68  —  a?  must  be  subtracted  from  84.  Observe  that  68  —  x 
is  not  so  great  as  68  by  x.  Therefore  if  I  subtract  68  from  84, 
I  shall  subtract  too  much  by  the  quantity  x,  and  I  must  add  x 
to  obtain  the  true  result. 

Then  we  have  84  —  68  +  a;  for  the  difference  between  84 
and  68  —  x. 

The  excess  of  40  above  the  less  is  40  —  x,  and  3  times  this 
is  120  — 3  a?. 

By  the  conditions,         84  —  68  -f-  a;  =  120  —  3a? 

Transposing  and  uniting,  4  a?  =  104 

Dividing  by  4,  a?  =:    26  =  less. 

68  —  26  =    42  —  greater. 

JVote.  In  this  question  68  —  x  was  subtracted  from  84.  In- 
stead of  a?,  now  put  its  value,  68  —  26.  Now  68  —  26  =:  42, 
that  is,  the  number  to  be  subtracted  from  84  is  42,  and  the  an- 
swer must  be  42.  When  68  is  subtracted  from  84,  the  result 
is  1 6,  which  is  too  small  by  26,  the  value  of  a? ;  to  this  it  is  ne- 
cessary to  add  26,  and  it  makes  42,  the  true  result,  84  —  68  -f- 
26  =r  42.  This  shows  that  we  did  right  in  adding  x  after  sub- 
tracting 68.     This  will  always  be  found  true.     Therefore, 


VI.  Equations.  35 

when  any  of  the  quantities  to  be  subtracted  have  the  sign  —  before 
them,  they  must  be  changed  to  -|-  in  subtracting,  and  those  which 
have  -{-  must  be  changed  to  — . 

2.  A  gentleman  hired  a  labourer  for  20  days  on  condition 
that,  for  every  day  he  worked,  he  should  receive  7s.,  but  for 
every  day  he  v^^as  idle,  he  should  forfeit  3s.  At  the  end  of  the 
time  agreed  on  he  received  80  shillings.  How  many  days  did 
he  work,  and  how  many  days  was  he  idle  ? 

Let  X  —  the  number  of  days  he  worked. 
Then  20  —  a:  =  the  number  of  days  he  was  idle. 
X  days,  at  7s.  a  day,  would  come  iol  x  shillings. 
20  —  x,  at  3s.  per  day,  would' De  60  —  3^  shillings.     This 
must  be  taken  out  of  7  x. 

By  the  above  rule  60  —  3  x,  subtra^ed  from  7  x,  leaves  7  x 
—  60  -f-  3  a:  ;  for  60  is  too  much  to  bJTsubtracted  by  3  x. 
By  the  conditions,    s  W 

7  a:— 60  4-3  a^-^  80. 
Transposing  and  uniting, 

10x=  140. 
Dividing  by  10,  a:  i=    14  =  days  he  worked. 

20  —  X  =1    6    ■=.  days  he  was  idle. 

3.  Two  men,  A  and  B,  commenced  trade  ;  A  had  twice  as 
much  money  as  B  ;  A  gained  $50,  and  B  lost  $90,  then  the 
difference  between  A's  and  B's  money  was  equal  to  three  tjmes 
what  B  then  had.     How  much  did  each  commence  with  f 

4.  Two  men,  A  and  B,  played  together  ;  when  they  com- 
menced they  had  $20  between  them,  after  a  certain  number 
of  games,  A  had  won  $6,  then  the  excess  of  A's  money  above 
B's  was  equal  to  |  of  B's  money.  How  much  had  each  when 
they  commenced  ^ 

5.  Divide  the  number  54  into  two  such  parts  that  the  less 
subtracted  from  the  greater,  shall  be  equal  to  the  greater  sub- 
tracted from  three  times  the  less.     What  are  the  parts  f 

6.  It  is  required  to  divide  the  number  204  into  two  such 
parts,  that  |  of  the  less  being  subtracted  from  the  greater,  the 
remainder  will  be  equal  to  f  of  the  greater  subtracted  from 
four  times  the  less. 

Let  X  =  greater  part. 

Then  204  —  a;  =  the  less  part. 


m 


36  Algebra.  VI. 


|ofthelessisl21-:iH. 


By  the  conditions, 

5  7 


Multiplying  by  5, 

5  a:  —  408  +  2  a?  =  4080— 20  a: —J-^. 

7 

Multiplying  by  7, 

35  a?-— 2856  +  14a:  =  28560—  140  a:—  15  a?. 
Transposing  and  uniting, 

204  a:  =31416 
a:  =  154 
204 — a:  =  50 
Let  X  denote  the  less  number,  and  solve  the  question  again. 
JVote.     Observe,  that  after  multiplying  by  5  in  the   above 
example,  the  signs  of  both  terms  of  the  numerator  were  chang- 
ed, that  of  408  to  — ,  and  that  of  2  a:  to  -|-  j  this  was  done  be- 
cause it  was  not  required  to  subtract  so  much  as  408  by  2  a?. 
The  change  of  signs  could  not  be  made  before  multiplying  by 
5,  because  the  sign  —  before  the  fraction  showed  that  the 
whole  fraction  was  to  be  subtracted.     If  the  signs  of  the  frac- 
tion had  been  changed  at  first,  it  would  have  been  necessary 
to  put  the  sign  -|-  before  the  fraction.     This  requires  particu- 
lar attention,  because  it  is  of  great  importance,  and  there  is 
danger  of  forgetting  it. 

7.  A  man  bought  a  horse  and  chaise  for  $341.  Now  iff  of 
the  price  of  the  horse  be  subtracted  from  twice  the  price  of  the 
chaise,  the  remainder  will  be  the  same  as  if  ^  of  the  price  of 
the  chaise  be  subtracted  from  three  times  the  price  of  the  horse. 
Required  the  price  of  each. 

8.  Two  men,  A  and  B,  were  playing  at  cards  ;  when  they 
began,  A  had  only  |  as  much  money  as  B.  A  won  of  B  $23  ; 
then  I  of  B's  money,  subtracted  from  A's,  would  leave  one 
half  of  what  A  had  at  first.  How  much  had  each  when  they 
began  ? 

9.  A  man  has  a  horse  and  chaise.  The  horse  is  worth  .$44 
less  than  the  chaise.  If  4  of  the  value  of  the  horse  be  sub- 
tracted from  the  value  of  the  chaise,  the  remainder  will  be  the 
same  as  if  from  the  value  of  the  horse  you  subtract  |  of  the  ex- 


VII.  Equations.  37 

cess  of  the  value  of  the  horse  above  84  dollars.     What  is  the 
value  of  the  horse  ? 

VII.  The  examples  in  this  article  are  intended  to  exercise 
the  learner  in  putting  questions  into  equation.  They  require 
no  operations  which  have  not  already  been  explained.  It 
was  remarked,  that  no  rule  could  be  given  for  putting  ques- 
tions into  equation,  but  there  is  a  precept  which  may  be  very 
useful. 

Take  the  unknown  quantity,  and  perform  the  same  operations  on 
it,  that  it  would  be  necessary  to  perform  on  the  answer  to  see  if  it 
was  right.      When  this  is  done  the  question  is  in  equation. 

1.  A  and  B,  being  at  play,  severally  cut  packs  of  cards  so  as 
to  take  off  more  than  they  left.  Now  it  happened  that  A  cut 
off  twice  as  many  as  B  left,  and  B  cut  off  seven  times  as  many 
as  A  left.     How  were  the  cards  cut  f 

Let       X  =  the  number  B  left. 

Then  2  a?  =  the  number  A  cut  off. 

52  —    x=:  the  number  B  cut  off. 

52  —  2  ^  =  the  number  A  left. 
By  the  conditions,  7  times  52  — 2x  are  equal  to  52  —  x. 
364  — 14a:  =52  — a;. 

Take  the  numbers  of  the  answer  and  endeavour  to  prove  that 
they  are  right,  and  you  will  see  that  you  take  the  same  course 
as  above. 

2.  A  man,  at  a  card  party,  betted  3s.  to  2  on  every  deal. 
After  twenty  deals  he  had  won  5  shillings.  At  how  many  deals 
did  he  win  ? 

Let  X  =:  the  number  of  deals  he  won. 

Then  20  —  a?  =  the  number  of  deals  he  lost. 

Every  time  he  won,  he  won  2  shillings  ;  that  will  be  2  a? 
shillings. 

Every  loss  was  3  shillings  ;  that  will  be  3  times  20  —  x,  or 
60— 3a:. 

The  loss  must  be  taken  from  the  gain,  and  he  will  have  5 
shiUings  left. 

2a:  — 60-f-3a?  =  5. 

3.  W^hat  two  numbers  are  to  each  other  as  2  to  3  ;  to  each 
of  which,  if  4  be  added,  the  sums  will  be  as  5  to  7. 

4 


38  Algehra,  VIL 

Let  X  =■  the  first  number. 

Then   — =  the  second. 
2 

Adding  4  to  each,  they  become  a?  -f*  4,  and  —  +  4. 

The  first  is  now  f  of  the  second,  or  the  second  is  J  of  the 
first. 

6  2""*" 

4.  A  sum  of  money  was  divided  between  two  persons,  A  and 
B,  so  that  the  share  of  A  was  to  that  of  B  as  5  to  3.  Now  A's 
share  exceeded  f  of  the  whole  sum  by  $50.  What  was  the 
share  of  each  person  ? 

Let  X  =.  A's  share. 

Then  ?f  z=  B's  share. 

5 

x4-  —  =  whole  sum. 
5  , 

5  ^f  ^  I    3a:.   5a?  ,   15a?    ^^  5a?  ,    a? 

foi  a?4-  — is  —  + ,  or  — +  — 

^5        9^  45'         9^3 


By  the  conditions, 


a?  =  ^-f-^  -I-  50. 
9^3^ 


5.  The  joint  stock  of  two  partners,  whose  particular  shares 
differed  by  48  dollars,  was  to  the  lesser  as  14  to  5.  Required 
the  shares. 

6.  Four  men  bought  an  ox  for  $43,  and  agreed  that  those, 
who  had  the  hind  quarters,  should  pay  J  cent  per  pound  more 
than  those,  who  had  the  fore  quarters.  A  and  B  had  the  hind 
quarters,  C  and  D  the  fore  quarters.  A's  quarter  weighed  153 
lb.,  B's  163  lb.,  C's  167  lb.,  and  D's  165  lb.  What  was 
each  per  lb.,  and  what  did  each  man  pay  ? 

7.  A  certain  person  has  two  silver  cups,  and  only  one  cover 
for  both.  The  first  cup  weighs  12  oz.  If  the  first  cup  be  co- 
vered it  weighs  twice  as  much  as  the  other  cup,  but  if  the  se- 
cond be  covered  it  weighs  three  times  as  much  as  the  first. 
What  is  the  weight  of  the  cover, and  of  the  second  cup  ? 


VII.  Equations.  39 

Let  X  =  weight  of  the  cover. 

Then     12  +  a?  =  weight  of  the  first  cup  covered. 

And       6  +-^=:  weight  of  the  second  cup,  &c. 

8.  Some  persons  agreed  to  give  6d.  each  to  a  waterman  for 
carrying  them  from  London  to  Gravesend  ;  but  with  this  con- 
dition, that  for  every  other  person  taken  in  by  the  way,  three 
pence  should  be  abated  in  their  joint  fare.  Now  the  waterman 
took  in  three  more  than  a  fourth  part  of  the  number  of  the  first 
passengers,  in  consideration  of  which  he  took  of  them  but  5d. 
each.     How  many  persons  were  there  at  first  f 

Let  X  =:  the  number  of  passengers  at  first. 

Then    —  +  3  =  the  number  taken  in,  &c. 

4 

9.  Four  places  are  situated  in  the  order  of  the  four  letters, 
A,  B,  C,  D.  The  distance  from  A  to  D  is  1 34  miles,  the  dis- 
tance from  A  to  B  is  to  the  distance  from  C  to  D,  as  3  to  2,  and 
one  fourth  of  the  distance  from  A  to*B,  added  to  half  the  dis- 
tance from  C  to  D,  is  three  times  the  distance  from  B  to  C. 
What  are  the  respective  distances  ? 

10.  A  field  of  wheat  and  oats,  which  contained  20  acres, 
was  put  out  to  a  labourer  to  reap  for  $20  ;  the  wheat  at  $1.20 
and  the  oats  $0.95  per  acre.  Now  the  labourer  falling  ill  reap- 
ed only  the  wheat.  How  much  money  ought  he  to  receive  ac- 
cording to  the  bargain  ? 

1 1 .  Three  men,  A,  B,  and  C,  entered  into  partnership  ;  A 
paid  in  as  much  as  B  and  one  third  of  C  ;  B  paid  as  much  as 
C  and  one  third  of  A  ;  and  C  paid  in  $10  and  one  third  of  A. 
What  did  each  pay  in  ? 

Let  X  =z  the  sum  A  contributed. 

Then        ^  -f- 10  =       "       C 
3  ^  ' 

and  -^ -f  10 -f -^  =:       "       B        »    &c. 

3  ^       ^3 

12.  A  gentleman  gave  in  charity  £46  ;  a  part  of  it  in  equal 
portions  to  5  poor  men,  and  the  rest  in  equal  portions  to  7  poor 
women.  Now  the  share  of  a  man  and  a  woman  together 
amounted  to  £8.  What  was  given  to  the  men,  and  what  to 
the  women  ? 


40  .  Algehra.  VII. 

Let  X  =  the  sum  a  man  received. 

Then  8  —  a:  =  the  sum  a  woman  received,  &c. 

13.  Suppose  that  for  every  10  sheep  a  farmer  kept,  he  should 
plough  an  acre  of  land,  and  should  be  allowed  an  acre  of  pas- 
ture for  every  4  sheep.  How  many  sheep  may  that  person 
keep  who  farms  700  acres  ^ 

Let  X  =  the  whole  number  of  sheep. 

The  number  of  acres  ploughed  will  be  j\  of  the  number  of 
sheep  ;  and  the  number  of  acres  of  the  pasture  will  be  J  of  the 
number  of  sheep  ;  both  these  added  together  must  be  the  whole 
number  of  acres,  &c. 

14.  A,  B,  and  C  make  a  joint  stock  ;  A  puts  in  $70  more 
than  B,  and  $90  less  than  C  ;  and  the  sum  of  the  shares  of  A 
and  B  is  I  of  the  sum  of  the  shares  of  B  and  C.  What  did 
each  put  in  ^ 

Let  X  =  the  sum  that  B  put  in,  &-c. 

15.  Divide  the  number  85  into  two  such  parts  that  if  the 
greater  be  increased  by  7  and  the  less  be  diminished  by  8,  they 
will  be  to  each  other  in  the  proportion  of  5  to  2. 

16.  It  is  required  to  divide  the  number  67  into  two  such  parts 
that  the  difference  between  the  greater  and  75  may  be  to  the 
excess  of  the  less  over  12  in  the  proportion  of  8  to  3. 

17.  A  man  bought  12  lemons  and  a  pound  of  sugar  for  56 
cents,  afterwards  he  bought  18  lemons  and  a  pound  of  sugar  at 
the  same  rate  for  74  cents.  What  was  the  price  of  the  sugar, 
and  of  a  lemon  ^ 

Let  X  =r  the  price  of  the  sugar. 

Then     56  —  a;  =  the  price  of  12  lemons. 

And       — m?  =  the  price  of  1  lemon. 
12  ^ 

In  the  same  manner, 

— m^  =  the  price  of  a  lemon. 

18  ^ 

Hence  ^lz:f  =  Z±ZL^,&c. 
12  18 

1 8.  A  man  bought  5  oranges  and  7  lemons  for  58  cents  ;  af- 
terwards he  bought  13  oranges  and  6  lemons  at  the  same  rate 
for  102  cents.  What  was  the  price  of  an  orange,  and  of  a 
lemon  ? 


VII.  Equations.                                         41 

Let  oc  =  the  price  of  an  orange. 

Then  — -  =  the  price  of  a  lemon  by  the  first  condi- 
tion, &c. 

19.  A  footman,  who  contracted  for  $72  a  year  and  a  Hvery 
suit,  was  turned  away  at  the  end  of  7  months,  and  received 
only  $32  and  the  livery.     What  was  the  value  of  the  livery  ? 

20.  A  landlord  let  his  farm  for  £10  a  year  in  money  and  a 
certain  number  of  bushels  of  corn.  When  corn  sold  at  10s.  a 
bushel,  he  received  at  the  rate  of  10s.  an  acre  for  his  land;  but 
when  it  sold  for  13s.  6d.  a  bushel,  he  received  13s.  an  acre. 
How  many  bushels  of  corn  did  he  receive  f 

Let  cc  =  the  number  of  bushels. 

Then      10  a;  -|-  200  =  the  year's  rent  in  shillings  ; 

10  a? -f  200  _.  -p   I   20  =  the  number  of  acres. 
10 
27  a:  -[-  400  =  the  year's  rent  at  the  second  rate  in  six- 
pences. 

It z=:  the  number  of  acres,  which  must  be  equal  to 

26  >  M 

the  other,  &c. 

21.  A  man  commenced  trade  with  a  certain  sum  of  money, 
which  he  improved  so  well,  that  at  the  year's  end  he  found  he 
had  doubled  his  first  stock  wanting  $1000  ;  and  so  he  went  on 
every  year  doubling  the  last  year's  stock  wanting  $1000 ;  at 
the  end  of  the  third  year  he  found  that  he  had  just  three  times 
as  much  money  as  he  commenced  with.  What  was  his  first 
stock  ? 

22.  A  man,  having  a  certain  sum  of  money,  went  to  a  tavern, 
where  he  borrowed  as  much  money  as  he  then  had,  and  then 
spent  a  shilling  ;  with  the  remainder  he  went  to  another  tavern, 
where  he  borrowed  as  much  as  he  then  had,  and  then  spent  a 
shilling,  and  so  he  went  to  a  third  and  a  fourth  tavern,  borrow- 
ing and  spending  as  before ;  after  which  he  had  nothing  left. 
How  much  money  had  he  at  first  ? 

23.  It  is  required  to  divide  the  number  60  into  two  such 
parts,  that  one  seventh  of  the  one  may  be  equal  to  one  eighth 
of  the  other. 

4* 


42  Algebra.  VII. 

24.  It  is  required  to  divide  the  number  85  into  two  such 
parts  that  f  of  the  one  added  to  ^  of  the  other  may  make  60. 

25.  It  is  required  to  divide  the  number  100  into  two*  such 
parts,  that  if  one  third  of  one  part  be  subtracted  from  one 
fourth  of  the  other,  the  remainder  may  be  11. 

26.  It  is  required  to  divide  the  number  48  into  two  such 
parts,  that  one  part  may  be  three  times  as  much  above  20,  as 
the  other  wants  of  20. 

27.  A  man  distributed 20  shillings  among  20  people,  giving 
6  pence  apiece  to  some,  and  16  pence  apiece  to  the  rest.  What 
number  of  persons  were  there  of  each  kind  .? 

28.  A  man  paid  £100  with  208  pieces  of  money,  a  part  gui- 
neas at  21s.  each,  and  a  part  crowns  at  5s.  each.  How  many 
pieces  were  there  of  each  sort  ? 

29.  A  countryman  had  two  flocks  of  sheep,  the  smaller 
consisting  entirely  of  ewes,  each  of  which  brought  him  2 
lambs.  On  counting  them  he  found  that  the  number  of 
lambs  was  equal  to  the  difference  between  the  two  flocks.  If 
all  his  sheep  had  been  ewes,  and  brought  forth  three  lambs 
apiece,  his  stock  would  have  been  432.  Required  the  number 
in  each  flock. 

Let  X  =1  the  number  in  the  less. 

Then       2x  =z  the  number  of  lambs. 

3  0?  =  the  number  in  the  larger. 

4  a?  =  the  number  in  both,  &c. 

30.  When  the  price  of  a  bushel  of  barley  wanted  but  3d.  to 
be  to  the  price  of  a  bushel  of  oats  as  8  to  5,  four  bushels  of  bar- 
ley and  7s.  6d.  in  money  were  given  for  nine  bushels  of  oats. 
What  was  the  price  of  a  bushel  of  each  ? 

Let  X  =:  the  price  of  a  bushel  of  oats  in  pence. 

8^ 
Then     —  —  3  ==  the  price  of  a  bushel  of  barley,  &c. 

31.  A  market-woman  bought  a  certain  number  of  eggs  at 
the  rate  of  2  for  a  cent,  and  as  many  at  3  for  a  cent,  and  sold 
them  out  at  the  rate  of  5  for  two  cents  ;  after  which  she  ob- 
served, that  she  had  lost  four  cents  by  them.  How  many  eggs 
of  each  sort  had  she  ? 


VII.  Equations.  43 

Let  X  =  the  number  of  each  sort. 

Then        -^  =  the  price  of  a:  eggs  at  2  for  a  cent. 

And  __  =  the  price  of  a?  eggs  at  3  for  a  cent. 

o 

These  added  together  make  what  the  eggs  cost. 

The  whole  number  is  2  a? ;  these  at  5  for  two  cents  come  to 

If  cents. 
5 

By  the  conditions,  £.  +  ^  =  If  -f-  4. 

32.  A  cistern  has  two  fountains  to  fill  it ;  the  first  will  fill  it 
alone  in  7  hours,  and  the  second  in  5  hours.  In  what  time  will 
the  cistern  be  filled,  if  both  run  together  ? 

Let  X  =r  the  number  of  hours  required  to  fill  it. 
The  first  would  fill  \  of  it  in  an  hour,  and  the  second  would 
fill  i  of  it  in  an  hour. 

Both  together  then  would  fill  |  -f-  i  in  an  hour  ;  and  in  x 

hours  both  would  fill  ^  4-  f-  of  it.     But  by  the  conditions  it 

7        5  ^ 

was  to  be  filled  in  x  hours. 

Therefore,       f.  -|-  f-  =  1  cistern. 

7^5 

* 

33.  A  gentleman,  having  a  piece  of  work  to  do,  hired  two 
men  and  a  boy  to  do  it ;  one  man  could  do  it  alone  in  5  days, 
the  other  could  do  it  alone  in  8  days,  and  the  boy  could  do  it 
alone  in  10  days.  How  long  would  it  take  the  three  together 
to  do  it  ? 

34.  A  cistern,  into  which  the  water  runs  by  two  cocks,  A 
and  B,  will  be  filled  by  them  both  running  together  in  12  hours ; 
and  by  the  cock  A  alone  in  20  hours.  In  what  time  will  it  be 
filled  by  the  cock  B  alone  ? 

Let  X  =:  the  time  in  which  B  will  fill  it  alone.  Both  will 
fill  j\  of  it  in  an  hour,  A  alone  gV  <^f  it,  and  B  will  fill  j\ — -^\ 
of  it  in  an  hour,  &c. 

35.  A  man  and  his  wife  usually  drank  out  a  vessel  of  beer  in 
12  days  :  but  when  the  man  was  from  home  it  would  usually 
last  the  wife  alone  30  days.  In  how  many  days  would  the  man 
alone  drink  it  out  ? 


44  Algebra.  "V^L 

36.  The  hold  of  a  ship  contained  442  gallons  of  water. 
This  was  emptied  out  by  two  buckets,  the  greater  of  which, 
holding  twice  as  much  as  thie  other,  was  emptied  twice  in  three 
minutes,  but  the  less  three  times  in  two  minutes ;  and  the 
whole  time  of  emptying  was  12  minutes.  Required  the  size  of 
each. 

The  greater  was  emptied  8  times  in  the  12  minutes,  (fee. 

37.  Two  persons,  A  and  B,  have  the  same  income.  A  saves 
\  of  his  ;  but  B,  by  spending  £80  a  year  more  than  A,  at  the 
end  of  4  years  finds  himself  £220  in  debt.  What  did  each  re- 
ceive and  expend  annually  ? 

38.  After  paying  ^  of  my  money,  and  \  of  the  remain- 
der, I  had  72  guineas  left.     How  much  had  I  at  first  ? 

39.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  the 
guineas  at  21s.,  and  the  moidores  at  27s.  each  ;  the  number 
of  pieces  of  both  sorts  was  just  100.  How  many  were  therq 
of  each  ^ 

40.  It  is  required  to  divide  the  number  26  into  three  such 
parts,  that  if  the  first  be  multiplied  by  2,  the  second  by  3,  and 
third  by  4,  the  products  shall  all  be  equal. 

Let  X  zzz  the  first  part.     The  second  part  must  be  _,  and 

the  third  part  ^or^. 

*^        4       2  • 

41.  It  is  required  to  divide  the  number  54  ipto  three  such 
parts,  that  i  of  the  first,  J  of  the  second,  and  J  of  the  third,  may 
be  all  equal  to  each  other. 

Let  2x=z  the  first  part. 

Then  Sx=  the  second  part,  &c. 

42.  A  person  has  two  horses  and  a  saddle,  which  of  itself 
is  worth  £25.  Now  if  the  saddle  be  put  upon  the  back  of  the 
first  horse,  it  will  make  his  value  double  that  of  the  second ; 
but  if  it  be  put  upon  the  back  of  the  second,  it  will  make 
his  value  triple  that  of  the  first.  What  is  the  value  of  each 
horse  ? 

43.  A  man  has  two  horses  and  a  chaise,  which  is  worth 
$183.  Now  if  the  first  horse  be  harnessed  to  the  chaise,  the 
horse  and  chaise  together  will  be  worth  once  and  two  sevenths 
the  value  of  the  other  ;  but  the  other  horse  being  harnessed, 
the  horse  and  chaise  together  will  be  worth  once  and  five 


VIII.  Equations  vrith  two  UnknovmQuantities.  45 

eighths  the  value  of  the  first.     Required  the  value  of  each 
horse. 

Equations  with  two  Unknown  Quantities. 

VIIL  Many  examples  involve  two  or  more  unknown  quan- 
tities. In  fact,  many  of  the  examples  already  given  involve 
several  unknown  quantities,  but  they  were  such,  that  they 
could  all  be  derived  from  one.  When  it  is  necessary  to  use 
two  unknown  quantities  in  the  solution,  the  question  must  al- 
ways contain  two  conditions,  from  which  two  equations  may 
be  derived.  When  this  is  not  the  case  the  question  cannot  be 
solved. 

1.  A  boy  bought  2  apples  and  3  oranges  for  13  cents  ;  he 
afterwards  bought,  at  the  same  rate,  3  apples  and  5  oranges 
for  21  cents.  How  much  were  the  apples  and  oranges 
apiece  ? 

Let  X  =  the  price  of  an  orange, 

and  y  =z  the  price  of  an  apple. 

1.  Sx-\-2y—13, 

2.  5x  -{-3y=z21. 

Multiply  the  first  equation  by  3,  and  the  second  by  2, 

3.  9x-]-6y=S9 

4.  10x-{-6y  =  42. 

Subtract  the  first  from  the  second,  because  the  y's  being 
alike  in  each,  the  difference  between  the  numbers  39  and  42 
must  depend  upon  the  x^s. 

5.  X  :=S  cents,  the  price  of  an  orange. 
Putting  this  value  of  x  into  the  first  equation, 

6.  9-}-2yr=13 

7.  y  =^  cents,  the  price  of  an  apple. 
Proof.     2  apples  at  2  cents  each  come  to  4  cents,  and  3 

oranges  at  3  cents  come  to  9  cents.     9  +  4  =  13.     So  3  ap- 
ples and  5  oranges  come  to  21  cents. 

JVote.  In  this  example  I  observed,  that  the  coeflicient  of  y 
in  the  first  equation  is  2,  and  in  the  second,  the  coefficient  of 
y  is  3.  I  multiplied  the  whole  of  the  first  equation  by  3,  and 
the  whole  of  the  second  by  2  ;  this  formed  two  new  equations 
in  which  the  coefficients  of  y  are  alike.  If  the  first  equation 
had  been  multiplied  by  5  and  the  second  by  3,  the  coefficients 
of  X  would  have  been  alike,  anct  x  instead  of  y  would  have  been 


46  Algehra.  VIII. 

made  to  disappear  by  subtraction,  and  the  same  result  would 
have  been  finally  obtained.  It  is  evident,  that  the  coefficients 
of  either  of  the  unknown  quantities  may  always  be  rendered 
alike  in  the  two  equations,  by  multiplying  the  first  equation  by 
the  coefficient  which  the  quantity  that  you  wish  to  make  dis- 
appear has  in  the  second  equation  ;  and  the  second  equation 
by  the  coefficient  which  the  same  quantity  has  in  the  first  equa- 
tion. They  may  be  rendered  alike  more  easily,  when  they 
have  a  conmion  multiple  less  than  their  product. 

2.  A  person  has  two  horses,  and  a  saddle  which  of  itself  is 
worth  £10  ;  if  the  first  horse  be  saddled,  he  will  be  worth  ^  as 
much  as  the  other,  but  if  the  second  horse  be  saddled,  he  \>  ill 
be  worth  j  as  much  as  the  first.  What  is  the  value  of  eacli 
horse  ? 

A  question  similar  to  this  has  already  been  solved  with  one 
unknown  quantity,  but  it  will  be  more  easily  solved  by  using 
two  of  them. 

Let  X  =  the  value  of  the  first  horse, 

and  y  =  the  value  of  the  second  horse. 

1.  By  the  conditions,  ^  =.x  -\-  10 

2.  "  ?f  =  y+10 

3.  By  transposition,  — ?  —  .j?=  10 

4.  "  ?5_y  =  10 

5        ^ 

Multiply  the  3d  by  7,  and  the  4th  by  5,  to  free  them  from 
denominators ; 

5.  — 7a:-|-6y  =  70 

6.  Sx — 5y=:50 

Multiply  the  5th  by  5  and  the  6th  by  6,  in  order  to  maKe 
the  coefficients  oiy  alike  in  the  two; 

7.  — 35a:+30y=  350 

8.  48cc— 30y=:  300 
Add  together  7th  and  8th, 

9.  48a:  — 35  a? -f30y  —  30y  =  350-1-300 

10.  Uniting  terms,  13  a:  =  650 

11.  0?=    50 


«■ 


VIII.  Equations  with  two  Unknown  Quantities,  47 

Putting  50,  the  value  of  a:,  into  the  5th, 

12  6y— 350=    70 

13.  61/ =  420 

14.  y—'^^ 

Ans.  The  first  is  worth  £50,  and  the  second  £70. 
JVote.     In  this  example  the  30  y  in  the  7th  equation  had  the 
sign  +,  and  in  the  8th  the  sign  —  before  it,  hence  it  was  ne- 
cessary to  add  the  two  equations  together  in  order  to  make  the 
y  disappear,  or  as  it  is  sometimes  called,  to  eliminate  y, 

3.  A  market-woman  sells  to  one  person,  3  quinces  and  4 
melons  for  25  cents,  and  to  another,  4  quinces  and  2  melons, 
at  the  same  rate,  for  20  cents.  How  much  are  the  quinces  and 
melons  apiece  ? 

4.  In  the  market  I  find  I  can  buy  5  bushels  of  barley  and  6 
bushels  of  oats  for  27s.,  and  of  the  same  grain  4  bushels  of 
barley  and  3  bushels  of  oats  for  18s.  What  is  the  price  of  each 
per  bushel  ? 

5.  My  shoemaker  sends  me  a  bill  of  $12  for  1  pair  of  boots 
and  3  pair  of  shoes.  Some  months  afterwards  he  sends  me  a 
bill  of  $20  for  3  pair  of  boots  and  1  pair  of  shoes.  What  are 
the  boots  and  shoes  a  pair  f 

6.  Three  yards  of  broadcloth  and  4  yards  of  taffeta  cost  57s., 
and  at  the  same  rate  5  yards  of  broadcloth  and  2  yards  of  taf- 
feta cost  81s.     What  is  the  price  of  a  yard  of  each  ? 

7.  A  man  employs  4  men  and  8  boys  to  labour  one  day,  and 
pays  them  40s. ;  the  next  day  he  hires,  at  the  same  wages,  7 
men  and  6  boys,  and  pays  them  50s.  What  are  the  daily  wages 
of  each  ? 

8.  A  vintner  sold  at  one  time  20  dozen  of  port  wine  and  30 
doz.  of  sherry,  and  for  the  whole  received  £120  ;  and  at  an- 
other time,  sold  30  doz.  of  port  and  25  doz.  of  sherry  at  the 
same  prices  as  before,  and  for  the  whole  received  £140.  What 
was  the  price  of  a  dozen  of  each  sort  of  wine  ? 

9.  A  gentleman  has  two  horses  and  one  chaise.  The  first 
horse  is  worth  $180.  If  the  first  horse  be  harnessed  to  the 
chaise,  they  will  together  be  worth  twice  as  much  as  the  se- 
cond horse  ;  but  if  the  second  be  harnessed,  the  horse  and 
chaise  will  be  worth  twice  and  one  half  the  value  of  the 
first.    What  is  the  value  of  the  second  horse,  and   of  the 

.  chaise  ? 


48  Algebra.  '     A. 

10.  Two  men,  driving  their  sheep  to  market,  A  says  to  B, 
give  me  one  of  your  sheep  and  I  shall  have  as  many  as  .you  ;  B 
says  to  A,  give  me  one  of  your  sheep  and  I  shall  have  twice  as 
many  as  you.     How  many  had  each  ^ 

Let  oc  =  the  number  A  had, 

And  y  =  the  number  B  had. 

If  B  gives  A  one,  their  numbers  will  be 

X  -\-\  andy —  1. 
If  A  gives  B  one,  their  numbers  will  be  ,, 

X  —  1  and  y  +  1,  (fee. 

11.  If  A  gives  B  $5  of  his  money,  B  will  have  twice  as 
much  as  A  has  left ;  but  if  B  gives  A  $5  of  his  money,  A 
will  have  three  times  as  much  as  B  has  left.  How  much  has 
each  ^ 

12.  A  man  bought  a  quantity  of  rye   and  wheat  for  £6,  the ' 
rye  at  4s.  and  the  wheat  at  6s.  per  bushel.     He  afterwards 
sold  1  of  his  rye  and  |  of  his  wheat  at  the  same  rate  for  £2. 
17s.     How  many  bushels  were  there  of  each  ? 

13.  A  man  bought  a  cask  of  wine,  and  another  of  gin  for 
$210  ;  the  wine  at  $1.50  a  gallon,  and  the  gin  at  $0.50  a  gal- 
lon. He  afterwards  sold  |  of  his  wine,  and  ^  of  his  gin  for 
$150,  which  was  $15  more  than  it  cost  him.  How  many  gal- 
lons were  there  in  each  cask  f 

14.  A  countryman,  driving  a  flock  of  geese  and  turkeys  to 
market,  in  order  to  distinguish  his  own  from  any  he  might  meet 
with  on  the  road,  pulled  three  feathers  out  of  the  tail  of  each 
turkey,  and  one  out  of  the  tail  of  each  goose,  and  found  that 
the  number  of  turkeys'  feathers  exceeded  twice  those  of  the 
geese  by  15.  Having  bought  10  geese  and  sold  15  turkeys  by 
the  way,  he  was  surprised  to  find  that  the  number  of  geese  ex- 
ceeded the  number  of  turkeys  in  the  proportion  of  7  to  3.  Re- 
quired the  number  of  each  at  first. 

Let            X  =  the  number  of  turkeys, 
and            y  =  the  number  of  geese. 
1 3x=2y  +  15 

2 y  +  io^^^-lQ^ 

^^  3 

3.  Freeing  the  2d  from  fractions,   3y-f-30  =  7a: — 105 

Instead  of  the  method  employed  above  for  eliminating  one 
of  the  unknown  quantities,  we  may  find  the  value  of  one  of 
them  in  one  equation,  as  if  the  other  were  known ;  and  then 


VIII.  Eqtutiions  with  two  Unknown  Quantities,  49 

this  value  may  be  substituted  in  the  other,  and  an  equation  will 
be  obtained,  containing  only  one  unknown  quantity,  which 
may  be  solved  the  usual  way. 

4.  Divide  the  first  by  3,  a?  =  ^^"^^^ 

*5 

5.  Multiply  the  4th  by  7,        lxz=z  lli(.±i?5 

o 
Substitute  this  value  of  7  ^  in  the  3d, 

^^  3 

7.  Multiply  by  3,  9  y+  90  =  14  y  -f  105  —  315 

8.  Transposing  &.  uniting,     300  =  5  y 

The  value  of  x  may  be  found  by  substituting  60  for  y  in 
the  4th, 

9.  ^  :.  =  i!2JLL?=45. 

,ikj*#'  Jlns.  45  turkeys,  and  60  geese. 

Let  the  learner  go  back  and  solve,  in  this  manner,  the  pre- 
ceding examples  in  this  Art.  Sometimes  one  method  is  pre- 
ferable and  sometimes  the  other. 

15.  A  person  expends  $1  in  apples  and  pears,  buying  his 
apples  at  3  for  a  cent,  and  his  pears  at  2  cents  apiece  ;  after- 
wards he  accommodates  his  neighbour  with  J  of  his  apples 
and  J  of  his  pears  for  30  cents.  How  many  of  each  did  he 
buy  ? 

Let  X  =  the  number  of  apples. 

And  y  ==  the  number  of  pears. 

Then         — =  the  price  of  the  apples. 
o 

And        2  y  =  the  price  of  the  pears,  &c. 

16.  A  market-woman  bought  eggs,  some  at  the  rate  of  2 
for  a  cent,  and  some  at  the  rate  of  3  for  two  cents,  to  the  amount 
of  65  cents  ;  she  afterwards  sold  them  all  for  120  cents,  and 
thereby  gained  one  half  cent  on  each  egg.  How  many  of  each 
kind  did  she  buy  ? 

17.  It  is  required  to  find  two  numbers  such,  that  if  J  of  the 
first  be  added  to  the  second,  the  sum  will  be  30,  and  if  J  of 
the  second  be  added  to  the  first,  the  sum  will  be  30. 

5 


50  Algebra,  VIII. 

18.  It  is  required  to  find  two  numbers  such,  that  |  of  the 
first  and  f  of  the  second  added  together  will  make  12,  and  if 
the  first  be  divided  by  2  and  the  second  be  multiplied  by  3,  | 
of  their  sum  will  be  26. 

19.  Two  persons,  A  and  B,  talking  of  their  ages,  says  A  to 
B,  8  years  ago  I  was  three  times  as  old  as  you  were,  and  4 
years  hence  I  shall  be  only  twice  as  old  as  you.  Required 
their  present  ages. 

20.  There  is  a  certain  fishing  rod,  consisting  of  two  parts, 
the  upper  of  which  is  to  the  lower  as  5  to  7  ;  and  9  times  the 
upper  part,  together  with  13  tiiv  es  the  lower  part,  is  equal  to 
1 1  times  the  whole  rod  and  8  feet  over.  Required  the  length 
of  the  two  parts. 

21.  A  vintner  has  two  kinds  of  wine,  one  at  5s.  a  gallon,  and 
the  other  at  12s.  of  which  he  wishes  to  make  a  mixture  of  20 
gallons,  that  shall  be  wortH  8s.  a  gallon.  How  many  gallons 
of  each  sort  must  he  use  ? 

22.  A  vintner  has  2  casks  of  wine,  from  each  of  which  he 
draws  8  gallons ;  and  finds  that  the  number  of  gallons  remain- 
ing in  the  less,  is  to  that  in  the  greater  as  2  to  5.  He  then 
puts  1  gallon  of  water  into  the  less,  and  5  gallons  into  the 
greater,  and  then  the  quantities  are  in  the  proportion  of  5  to  13. 
What  quantity  did  each  contain  at  first  ? 

23.  A  farmer,  after  selling  13  sheep  and  5  cows,  found  that 
the  number  of  sheep  he  had  remaining,  was  to  that  of  his  cows 
in  the  proportion  of  4  to  3.  After  three  years  he  found  that 
he  had  57  more  sheep,  and  10  more  cows  than  he  had  at  first ; 
and  that  the  proportions  were  then  as  3  to  1 .  What  number 
of  each  had  he  at  first  ? 

24.  When  wheat  was  8  shillings  a  bushel,  and  rye  5  shil- 
lings, a  man  wished  to  fill  his  sack  with  a  mixture  of  wheat 
and  rye,  for  the  money  he  had  in  his  purse.  If  he  bought  15 
bushels  of  wheat,  and  laid  out  the  rest  of  his  money  in  rye,  he 
would  want  3  bushels  to  fill  his  sack ;  but  if  he  bought  1 5 
bushels  of  rye,  and  then  filled  his  sack  with  wheat,  he  would 
have  15  shillings  left.  How  much  of  each  must  he  purchase 
in  order  to  lay  out  his  money  and  fill  his  sacks  ? 

25.  A  grocer  had  2  casks  of  wine,  the  smaller  at  7s.  per  gal- 
lon, the  larger  at  10s.     The  whole  was  worth  $112.     When 


IX.  Equations,  Generalization.  51 

he  had  drawn  18  gals,  from  each,  he  mixed  the  remainder  to- 
gether and  added  3|  gals,  of  water,  and  the  mixture  was  worth 
8s.  per  gal.  How  many  gallons  of  each  sort  were  there  at 
first  ? 

Equations,  Generalization. 

IX.  In  the  examples  hitherto  proposed  a  numerical  result 
has  always  been  obtained.  The  solution  with  numbers  has 
been  performed  at  the  same  time  with  the  reasoning  ;  and 
when  the  work  was  finished,  no  traces  of  the  operations  re- 
mained in  the  result.  But  algebra  has  a  more  importani,  pur- 
pose. Pure  algebra  never  gives  a  numerical  result,  but  is  used 
to  trace  general  principles  and  to  form  rules.  In  order  to  pre- 
serve the  work  so  that  the  operations  may  appear  in  the  result, 
it  will  be  necessary  to  introduce  a  few  more  signs^ 

1.  It  is  required  to  divide  $500  between  two  men,  so  that 
one  of  them  may  have  three  times  as  much  as  the  other. 
Let  X  =z  the  less  part. 
The  equation  will  be       x  -\-Sx  =:  500 

4  ir  =  500 
X  =  125 
3  a?  =  375 
^ns.  One  part  is  $125,  and  the  other  $375. 
This  question  is  to  divide  500  into  two  such  parts,  that  one 
part  may  be  three  times  as  much  as  the  other.     It  is  evident 
that  the  process  will  be  the  same  for  any  other  number,  as 
for  500. 

Let  the  number  to  be  divided  be  represented  by  the  letter  a. 
This  will  stand  for  any  number. 

Then  the  question  will  be,  to  divide  any  number,  a,  into 
two  such  parts,  that  one  part  may  be  three  times  as  much  as 
the  other. 

The  equation  will  he   x-\-3x  z=z  a 

4x  =:  a 


X 

a 

3x 

__2a 

52  Algebra.  IX. 

The  work  is  now  preserved  in  the  result,  and  it  appears  that 
one  part  will  be  J  of  the  number  to  be  divided  ;  and  the  other, 
f  of  it.     This  is  a  rule  that  will  apply  to  any  number. 

Suppose  a  =.  500  as  in  the  example. 

Then    Ji  —  125  ;  and  ^  =  375. 
4  4 

Ans.  One  part  is  $125,  and  the  other  $375 ;  the  same  as 
above. 

Suppose  it  is  required  to  divide  $7532  in  the  same  propor- 
tions. 

Then     a  =  7532 ;  —  =  1883  ;  and  ii*  =  5649. 
4  '4 

Ans.  One  part  is  $1883,  and  the  other  is  $5649. 
2.  A  man  sold  some  apples,  some  pears,  and  some  oranges 
for  a  number  a  of  cents,  the  apples  at  two  cents  apiece,  the 
pears  at  three  cents  apiece,  and  the  oranges  at  five  cents 
apiece.  There  were  twice  as  many  pears  as  oranges,  and 
three  times  as  many  apples  as  pears.  How  many  were  there 
of each  ? 

Let  X  :=  the  number  of  oranges. 

Then       2  ^  =  the  number  of  pears. 
And         6  a:  =  the  number  of  apples. 
By  the  conditions,  12a?-|-6a:-f-5a:=:  a 

2Sa:=  a 

X  = —  =  No.  of  oranges. 

2xzJi^=:  «    of  pears.       . 

6  a?  =  _?  r=   "    of  apples. 
23  ^^ 

Suppose  a  =  184  cents,  then  ^V  of  184  =  8  =  the  number 
of  oranges  ;  2  X  8  ==  16  :=  the  number  of  pears  ;  and  6x8 
=  48  =:  the  number  of  apples.  This  is  easily  proved.  8 
oranges,  at  5  cents  apiece,  come  to  40  cents  ;  16  pears,  at  3 
cents  apiece,  come  to  48  cents  ;  and'  48  apples,  at  2  cents 
apiece,  come  to  96  cents  ; 

40  +  48  +  96=184. 
The  learner  may  be  curious  to  know,  how  it  is  possible  to 
make  the  examples  in  such  a  manner,  that  the  answer  may  al- 


IX.  Equations,  Generalization,  53 

ways  come  out  a  whole  number  when  it  is  wished  ;  for  if  the 
numbers  were  taken  at  random,  there  would  frequently  be  frac- 
tions in  the  result.  The  method  is  to  solve  it  first  with  a  letter, 
as  has  been  done  in  the  two  preceding  examples.  If  any  num- 
ber, which  is  divisible  by  4,  be  put  in  the  place  of  a,  in  the 
first  example,  the  answer  will  be  in  whole  numbers.  And 
if  any  number,  which  is  divisible  by  23,  be  put  in  the  place  of 
a,  in  the  second  example,  the  answer  will  be  in  whole  num- 
bers. 

Let  the  learner  now  generalize  the  examples  in  Art.  I.,  by 
substituting  a  letter  instead  of  the  number ;  and  after  the  re- 
sult is  obtained,  put  in  the  numbers  again,  and  see  if  the  an- 
swers agree.     Let  him  also  try  other  numbers. 

The  examples  in  Art.  II.  may  be  generalized  in  the  same 
manner. 

3.  A  man  being  asked  his  age,  answered,  that  if  its  half  and 
its  third  were  added  to  it,  the  sum  would  be  88.  Required 
his  age. 

Instead  of  88  put  a,  and  let  x  =.  the  number  required. 

X  4r —  +  —  =z  a 
11  a: 


11  x=:  6 a 
6  « 

Any  number  that  is  divisible  by  11,  being  put  in  the  place 
of  a,  will  give  an  answer  in  whole  numbers.  Let  a  =  88,  then 
j\  of  it  is  48,  agreeing  with  the  answer  in  Art.  II. 

In  the  course  of  the  solution  it  appears,  that  a  is  equal  to  y 
of  a;  ;  and  the  result  shows,  that  x  is  equal  to  j\  of  a.  That 
is,  the  value  of  x  is  found  by  multiplying  a  by  the  fraction  y 
inverted. 

4.  In  an  orchard  of  fruit-trees,  ^  of  them  bear  apples,  J  of 
them  cherries,  and  the  remainder,  which  is  a,  bear  peaches. 
How  many  trees  are  there  in  the  orchard  ^ 
5* 


54  Algebra.  Ct. 

Let    X  =  the  whole  number  of  trees. 
Then 


X 

-T* 

f^ 

ha 

12  X 
12 

=  i-  + 

3a: 
12 

4- « 

bx 
12 

=  a 

5a? 

X 

=  12a 
__  12tt 

5 

Any  number  that  is  divisible  by  5,  may  be  put  in  the  place 
of  a.     If  a  =:  15,  the  answer  is  36. 

Proof.  25  +  5^-1-15  =  36. 

3^4^ 

5.  The  8th  example  of  Art.  II.  is  solved  as  follows  : 
Instead  of  100  put  a,  and  let  x  =  the  whole  number  of 
geese. 


Then 

X  +  X  + 

f  +  2i  =  a 

Multiplying  by  2, 
By  transposition, 

5a:- 

■f-5  =  2a 
5a?  =  2a — 5 

^       2a  — 5.  ^^ 

X  =  — -_ ;  or 

o 

2a        5        2a 

X  zzz — — in  — 

5        5         5 

1 

Let 

a  =  100. 

Then 

^  _  2  X  100 
5 

5 

or 

,  ^  _  2  X  100 

—  1=40—1=39. 

Let  a  =:  135,  and  find  the  answer  in  the  same  way. 
The  answer  will  be  53. 
Proof.  53+53-1-  26 J  +  2 J  =  1 35. 

The  learner  may  now  generalize  the  examples  in  Art.  II. 
The  preceding  examples  admit  of  being  generalized  still 


IX.  Equations,  Generalization,  55 

more,  but  the  process  would  be  too  difficult  for  the  learner  at 
present.     The  following  question  admits  it  more  easily. 

6.  (Art.  III.  Exam.  1.)  Two  men,  A  and  B,  hired  a  pasture 
for  f  55,  and  A  was  to  pay  $13  more  than  B.  How  much  did 
each  pay  ? 

This  question  is,  to  divide  the  number  55  into  two  such 
parts,  that  one  may  exceed  the  other  by  13. 

*  Let  us  represent  55  by  a,  and  13  by  6.  The  question  now 
is  to  divide  the  nxHnber  a,  into  two  such  parts,  that  one  may 
exceed  the  other  by  the  number  b  ;  a  and  b  being  any  two 
numbers,  of  which  a  is  the  larger. 

Let  X  =  the  less  part. 

Then  a?  +  6  =  the  greater  part. 

And  X  -^x-^-b  =  u 

2x  4-&  =  ct 
By  transposition,  2x=:a  —  b 

Dividing  by  2,  x=l.-^  =  t=l 

When  a  number,  consisting  of  two  or  more  parts^  as  a  —  J,  is 
to  be  divided,  it  is  evident  that  all  the  terms  must  be  divided, 

as  ^  —  —   But  the  fractions  ~  and  — ,  having  a  common  de- 

2         2  2  2^ 

nominator,  one  numerator  may  be  subtracted  from  the  other. 

n  h  fi  — .  h 

Hence  —  —  —  is  the  same  as .    This  is  easily  seen  in 

2  2  2  ^ 

numbers.  See  below,  where  55  and  13  are  substituted  for  a 
and  b. 

Hence  it  appears,  that  the  less  part  is  found  by  subtracting  half 
of  the  excess  of  the  greater  above  the  less  from  half  the  number  to  be 
divided  ;  or  by  taking  half  the  difference  between  the  number  to  be 
divided  and  the  excess. 

The  greater  part  is  equal  to  x  +  b;  hence  if  6  be  added  to 

—  —  —  it  will  give  the  greater  part : 

*  Whenever  the  learner  finds  any  difficulty  in  comprehending  the  operations 
in  the  gentirai  solutions,  let  him  first  solve  the  questi(ms  with  the  numbers. 


66  .  Algebra.  IX. 

^  2         2  ^     ' 

or  07-4-0=:  —  —  —  4-  —  ; 

^  2         2^2 

or  ^  +  6  =  J^  +  l  =  fl±i. 

2         2  2 

TAe  greater  is  found  by  adding  half  the  excess  to  half  the  number 
to  be  divided  ;  or  by  talcing  half  the  sum  of  the  number  to  he  divided 
and  the  excess. 

In  the  above  example, 

A's  part  ..  55  +  15,  or  '^ '^  =  34. 
^  2^2  2 

B's  part  =  55—1?,  or  55rzi?  =  21. 

^  2        2  2 

Let  the  learner  generalize  this  question  by  making  x  =  the 
greater  part.     The  same  results  will  be  obtained. 

This  is  a  general  rule,  and  will  apply  to  all  questions  like  it, 
and  should  be  remembered,  for  it  is  frequently  useful. 

Let  the  learner  find  the  answers  to  the  2d,  3d,  and  7th  ex- 
amples of  Art.  III.  by  this  rule.  That  is,  by  putting  the  num- 
bers of  those  examples  in  the  place  of  a  and  b  in  the  formulas. 

It  is  easy  to  see  the  propriety  of  the  rule.     For  the  formula 

?:Zl5or55nii=l?,  shows,  that  if  the  $13  that  A  pays 

X>  ^  >i 

more  than  B,  be  taken  out,  the  remainder  is  to  be  paid  in  equal 

parts  ""by  them.     Also  the    formula  ^   '      or it — = — , 

^  ^  2  2  2 

shows,  that  if  B  were  to  pay  $13  more,  he  would  pay  as  much 
as  A,  and  the  rent  would  be  paid  in  equal  parts  by  them. 

7.  A  father,  who  has  three  sons  (Art.  III.  exam.  4),  leaves 
them  16000  crowns.  The  will  specifies,  that  the  eldest  shall 
have  2000  crowns  more  than  the  second,  and  that  the  second 
shall  have  1000  crowns  more  than  the  third.  What  is  the  share 
of  each  f 

Let  a  represent  the  whole  number  of  crowns,  b  what  the 
eldest  son's  share  exceeds  that  of  the  socond,  and  c  what  the 
share  of  the  second  son  exceeds  that  of  the  third. 

This  question  may  be  expressed  in  general  terms,  thus  :  To 
divide  a  given  number  a,  into  three  such  parts,  that  the  great- 


IX.  Equations,  Generalization.  57 

est  may  exceed  the  mean  by  a  given  number  6,  and  the  mean 
may  exceed  the  least  by  a  given  number  c 
Let  x=z  the  greatest. 

Then  x- —  6  =  the  mean. 

And         X  —  b  —  c  =  the  least. 
By  the  conditions, 

x-\'X  —  b -{- X — b  —  c=ict 
Sx — 2b  —  c=a 
By  transposition,  2x=  a-\-2b  -{-c 

Dividing  by  3,  cc  =  ^  +  ^l  +  ±. 

o  t>  «5 

Or  because  the  fractions  have  a  common  denominator, 
^_a^2b  +  c 
3 
This  is  the  formula  for  the  greatest  part.     The  mean  is  «  — 

b,  or  b  subtracted  from  -  4-  —  4-  £,  thus  ; 
3         3        3 

3^   3^3 

I        a   ,   2b   ,   c        Sb 
or  X  —  6=:--|-  —  4-  -  —  — , 

3        3  ^3        3 

^       3"*'3  3 

The  least  part  is  a?  —  b  —  c,  or  c,  subtracted  from 

a b    ,   c 

3       3"^3' 

7^ a ^    t^ 

^"""3       33       ^' 

^^       ^       I, a       b    .  c       Sc 

or       X  —  0  —  c  =  -  —  -  +  —  —  — » 
3       3/^3        3 

1 a b 2  c a  —  b  —  2  c 

""3       3        3"  3 


58  Algebra,  IX. 

The  greatest  part  is  "«_+?!+_£. 


The  mean        do. 
The  least         do. 


a  —  h  -f-  c 

3 
a  —  h  —  2c 


3 
The  eldest  son's  share,  by  the  first  formula,  is 

16000  +  2  X  2000  +  1000   ^^^^ 

J. 11 X =  7000  crowns. 

3 

The  other  shares  may  be  found  by  the  other  two  formulas. 

Let  the  learner  solve  this  question  by  making  x  equal  to  the 
less  part,  and  also  by  making  it  equal  to  the  mean. 

Exam.  5th,  Art.  III.  may  be  solved  by  this  formula.  Let 
the  learner  generalize  the  questions  in  Art.  IIL  as  far  as  to 
Exam.  16th. 

The  examples  in  Art.  L  may  be  generalized  still  farther. 

8.  A  man  bought  corn  at  4s.  (a)  per  bushel,  rye  at  6s.  [h) 
per  bushel,  and  wheat  at  8s.  (c)  per  bushel :  there  was  an  equal 
quantity  of  each  sort.  The  whole  came  to  90s.  {d).  How  ma- 
ny bushels  were  there  of  each  ^ 

It  will  readily  be  perceived  that  it  is  impossible  actually  to 
perform  the  operations  of  addition,  subtraction,  &c.  on  letters  ; 
but  it  is  easy  to  represent  these  operations.  We  however  fre- 
quently speak  of  adding,  subtracting,  multiplying,  and  dividing 
algebraic  quantities,  by  which  we  mean,  representing  these 
operations.  We  have  seen  that  to  express  3  times  a?  or  3  times 
a  we  write  3  a:,  3  a,  that  is,  x  ox  a  multiplied  by  3.  In  the  same 
manner,  if  we  wish  to  express  a  times  a?,  that  is,  x  multiplied 
by  a,  we  write  a  x  ;  and  if  we  wish  farther  to  express  that  a  x 
(that  is,  a  times  x)  is  to  be  multiplied  by  5,  we  write  ah  x, 

*Let  X  =:  the  number  of  bushels  of  each. 

Then         ax  =:  the  price  of  the  com. 
bx  =  the  price  of  the  rye. 

And  ex  =  the  price  of  the  wheat. 

ax-{-bx-\-cx=:d. 

Here  x  is  taken  a  times,  and  b  times,  and  c  times,  that  is, 
{a  -{-b  -|-c)  times.   This  may  be  expressed  thus,  (a  -f-  6  -f-  c)  cr, 

•*  Let  the  learner  perform  this  example  first  by  the  numbers. 


IX.  Equations i  Generalization,  &^ 

enclosing  the  three  coefficients  connected  by  their  signs  in  a 
parenthesis. 

This  will  be  plain  if  we  put  it  in  numbers. 
4  a:  -f-  6  a?  -{-  8  a:  is  the  same  as  (4  +  6  +  8)  a?,  that  is,  18  x. 

\a -{- h  -{' c)  X  z=  d 
If  we  had  lSx:=d 

we  should  divide  by  18,  a:  =  — 

^      '  18 

[n  the  same  manner  divide  by  (a  +  6  +  c), 

X  =z 

a  -\-  b  -\-c 
Particular  Am.  5  bushels. 
This  general  formula  is  expressed  in  words  as  follows  :  Di- 
vide the  price  of  the  whole  by  the  price  of  a  bushel  of  each 
sort  added  together,  and  it  will  give  the  number  of  bushels  of 
each  sort. 

9.  A  father  dying  left  $25000  (or  a)  to  be  divided  between 
his  wife,  son,  and  daughter ;  his  son  was  to  have  3  (or  b)  times 
as  much  as  the  daughter,  and  the  wife  2  (or  c)  times  as  much 
as  the  son.     What  was  the  share  of  each  ? 
Let  X  ==  the  share  of  the  daughter. 

Then       Sxorb  x  =  the  share  of  the  son. 
And         6xoTbcx=z  the  share  of  the  wife, 
a?  +  3a? -1-6 a?  ==25000 
X  -\-bx  -{-bcx  =^  a 
(14.3  +  6)0?=  10  a:  =  25000 
{l-i'b-{-bc)x  =  a 

a:  =  ^^-2-  =  2500 
10 


X  = 


l-\-b  +  bc 


In  this  example  observe  that  x  is  taken  1  time,  and  b  times, 
and  be  times.  When  a  letter  is  vn-itten  without  a  coefficient, 
it  is  always  understood  to  have  1  for  its  coefficient ;  thus  x  is 
the  same  as  1  x. 

Having  found  the  share  of  the  daughter,  it  is  easy  to  find  the 
shares  of  the  other  two. 


«0  Algebra,  IX. 

The  son's  share  is  3  a?  =  7500,  or  6  a:  =         ^  ^ 


1  -|-6  +  6c 
a6c 


The  wife's  do.  is  6  a?  =  15000, oxhcxz=: 

l+6  +  6c 

The  learner  may  now  generalize  some  of  the  examples  in 
Art.  I.  in  this  manner. 

10.  A  gentleman,  distributing  some  money  among  some 
beggars,  found,  that  in  order  to  give  them  8  (or  a)  cents  apiece, 
he  should  want  5  (or  h)  cents  ;  he  therefore  gave  them  7  (or  c) 
cents,  and  he  had  4  (or  d)  cents  left.  How  many  beggars  were 
there  ^ 

Let  X  =  the  number  of  beggars. 

Then  8a?—  5    =  7  a? -f  4 

or  ax  —  h    =1  c  X  -\-  d 

8a?-- 7a?=  5    -j"  4  =  9 
a  X  —  ex  z=  b    -\-  d 
(8— 7)a?=:  9 
(a  —  c)  a?  =  6    -f-  c? 
x=.  9 

a  — c 

Particular  Ans.  9  beggars. 

General  Ans,  ^-±A, 
a  — c 

11.  There  is  a  cistern  which  is  supplied  by  two  pipes  ;  the 
first  will  fill  it  alone  in  7  (or  a)  hours,  the  second  will  fill  it 
alone  in  5  (or  b)  hours.  In  what  time  will  it  be  filled  if  both 
run  together  ? 

Let  a?  =:  the  number  of  hours  in  which  both  together  will 
fill  it. 

The  first  will  fill  4  or  -.  of  it  in  one  hour,  and  the  second  will 
a 

fill  I  or-  of  it  in  one  hour  ;  both  together  will  fill  4  -|-  j  or 
b 

-  +  -  of  it  in  one  hour.     In  x  hours  they  will  fill  x  times  as 
a      b 

much,  that  is, 


X     I     X     ^_    X     I     X 

--+--,  or  -  +  J-' 
1       b        a       b 


IX.  Equations,  Generalization.  '  61 

But  X  hours  is  the  whole  time,  therefore,  the  cistern  being  I, 

7^5  a^h 

Clearing  of  fractions, 

5a;  +  7a?  =  35       hx  -\'  ax=.ah 
Uniting  coefficients,         12a:  =  35       {b  -\'  a)  x=:ah 


X=:2\\  X  = 


ah 


a-^h 
Particular  Am,  2\\  hours. 

General  Am. 


Suppose  one  pipe  would  fill  the  cistern  in  8J  hours,  and  the 
other  in  4f  hours,  and  find  the  answer  by  the  general  formula. 

Am.  3yf  g  hours. 

12.  Suppose  it  were  required  to  make  a  rule  for  Fellowship. 
First  take  a  particular  case. 

Three  men,  conunencing  trade  together,  furnished  money  in 
the  following  proportions  ;  A  f  S  as  often  as  B  $5,  and  as  often 
as  C  $3.  They  gained  $800.  What  is  each  man's  share  of 
the  gain  ? 

It  is  evident  that  they  must  receive  in  the  proportion  of  the 
capital  that  they  respectively  furnished. 


Let 

X  : 

=  A's  share  of  the  gain. 

Then 

5a? 

"8"  ■ 

=  B's  share. 

And 

Sx_ 

8   ■ 

=  C's  share.* 

:W^ 

^8^8 
8  a? -i- 5a?  +  3a?  =  6400 
16  a?  =6400 

a?=    400  =  A's  share. 

^=:    250  =  B's  share. 

8 

£5  =    150  =  C's  share. 

8 

*  See  Art.  II.  Examp.  24  and  25. 
6 

62  Algebra,  IX. 

Now,  instead  of  8,  5,  and  3,  suppose  they  furnished  in  the 
proportion  of  w,  «,  and^;  and  let  the  whole  gain  be  a.. 

Let  X  —  A's  share  of  the  gain. 

Then       !L^  =  B's  share. 
m 

And         PJ^  =  C's  share. 
m 

Then  we  have 

,    nx   ,  px 

m         m 

mx -{-nx -{- px  zrzma 

{m'\-n  -\- p)  X  zizma 

ma  . ,     , 

A's  share. 


m-j-n  -\- p 


B's  share  is  ^—,  or  the  —  part  of =  A's  share. 

m  m  m  -\-  n  -j-p 

Since  a  fraction  is  divided  by  dividing  its  numerator,  the 

-  part  of ^^ ,  will  be  found  by  dividing  the  numerator 

m  m  -{-n  -{-p 

m  ahyin,     a  multiplied  by  w  is  m  a,  therefore,  m  a  divided  by 

m  is  a.     Hence  the   _  part  of is ,  and 

m  m  -{•  n  -\-  p       m  -{-  n-^p 

the    —  part  is  n  times  as  much,  that  is ,  which  is  B's 

m  m  +  n-{'p 

share. 

C's  share  is^,  or  the  ^  part  of — ,    which    is 

m  m  m  -\-  n  -\'  p 

pa 


-[-n+p 


A's  share  is "^ ;  B's  do.  -^— ;  and  C's  do. 

m  +  n  +  p  m  +  w-{-jp 


pa 


IX.  Equations,  Generalization.  63 

Hence  to  find  the  share  of  either,  multiply  the  whole  sum  to  be 
divided,  by  the  proportion  of  the  stock  which  he  furnished,  and  di- 
vide the  product  by  the  sum  of  their  proportions. 

The  propriety  of  this  rule  is  easily  seen.     For,  putting  in  the 

Q 

numbers  instead  of  the  letters,  A's  share  is or  rXr  of 

8-1-5  +  3      ^ 

$800,  B's  share  is or  /^  of  it,  and  C's  share  is 

8  +  54-3         ' 

3 

or  y\  of  it.     That  is,  the  sum  of  all  their  propor- 

8  +  5  +  3 

tions  is  16,  and  of  these  A  furnished  8  ;  B,  5  ;  and  C,  3. 

13.  Let  it  be  required  to  find  what  sum,  put  at  interest  at  a 
given  rate,  will  amount  to  a  given  sum  in  a  given  time  ;  that 
is,  to  find  a  rule,  by  which  the  principal  may  be  found,  when 
the  rate,  time,  and  amount  are  given. 

First  take  a  particular  case. 

A  man  lent  sonie  money  for  3  years,  interest  at  6  per  cent, 
and  received  for  interest  and  principal  $472.  What  was  the 
sum  lent  ? 

Let  X  =  the  sum  lent. 

Then  _^  ==  the  interest  for  1  year. 

100  ^ 

And  L^=         do.        for  3  years. 

100  ^ 

-i  Q 

And         X  + =  the  amount  for  3  years. 

'100  ^ 

IP 

Hence  we  have      x  + =  472 

100 

100  a?  +  18a;  =  47200 
118  0^  =  47200 

X  =  $400  =  The  sum  lent. 
It  is  a  custom  established  among  mathematicians  to  use  the 
first  letters  of  the  alphabet  for  known  quantities,  and  some  of 
the  last  letters  for  unknown  quantities.  It  is,  however,  fre- 
quently convenient  to  choose  letters,  that  are  the  initials  of  the 
words  for  which  they  stand,  whether  the  quantities  be  known 
or  unknown. 


64  Jllgebra,  IX. 

To  generalize  the  above  example, 

Let  p  1=  the  principal,  or  sum  lent. 

r  z=  the  rate  per  annum,  which  in  the  above  case 
is  yf  7  or  .06. 

and  t  z=z  the  time  for  which  it  was  lent, 

and  a  =  the  amount. 

Then       rp  =  the  interest  for  one  year, 

and        trp=         do.         for  t  years, 

andp  -\-trp^  the  amount. 
Hence  we  have        p  -\-  trp  =.a 
(l  +  tr)p  =  a 

p  =  -±-. 
I  -ytr 

That  is,  multiply  the  rate  by  the  time,  add  1  to  the  product,  and 
divide  the  amount  by  this,  and  it  tuill  give  the  principal. 

In  the  above  example  the  rate  is  .06,  which,  multiplied  by  3 
(the  time),  gives  .18,  and  one  added  to  this  makes  1.18  ;  472 
divided  by  1.18  gives  400,  as  before. 

Apply  this  rule  to  the  following  example. 

A  man  owes  $275,  due  two  years  and  three  months  hence, 
without  interest.  What  ought  he  to  pay  now,  supposing  money 
to  be  worth  4J  per  cent,  per  annum  ? 

N.  B.  2  years  and  3  months  is  2^  years. 

Ans.  |249yV/TV6- 

See  Arithmetic,  page  84. 

The  learner  may  now  make  rules  for  the  following  purposes  : 

14.  The  interest,  time,  and  rate  being  given,  to  find  the  prin- 
cipal. 

15.  The  amount,  time,  and  principal  being  given,  to  find  the 
rate. 


16.  The  amount,  principal,  and  rate  given,  to  find  the  time. 


17.  A  man  agreed  to  carry  20  (or  a)  earthen  vessels  to  a 
certain  place,  on  this  condition  ;  that  for  every  one  delivered 
safe  he  should  receive  8  (or  b)  cents,  and  for  every  one  he 
broke,  he  should  forfeit  12  (or  c)  cents  ;  he  received  100  (or  a') 
cents.     How  manv  did  he  break  ? 


IX.  Equations,  Generalization,  65 

Let  X  =  the  number  unbroken. 

Then  20  —  x  or  a  —  x  =:  the  number  broken. 

For  every  one  unbroken  he  was  to  receive  8  or  6  cents,  these 
will  amount  to  8  a?  or  6  a? ;  and  for  every  one  broken  he  was 
to  pay  back  12  or  c  cents,  these  will  amount  to  240  —  12  a? 
cents,  or  ac-—cx  ;  this  must  be  subtracted  from  the  former. 

240  —  12  X,  subtracted  from  8  a;,  is 

8a?  — 240  -f  12a?,  or  20  a?  —  240. 
Also  ca  —  ex  subtracted  from  b  x,isbx  —  c  a  -{-  c  x;  for  the 
quantity  ca  —  c  a:  is  not  so  large  as  c  a,  by  the  quantity  c  x, 
therefore  when  we  subtract  c  a  from  b  x,  we  subtract  too  much 
by  c  X,  and  in  order  to  obtain  a  correct  result,  it  is  necessary  to 
add  c  X. 

The  equation  is 

20a:  — 240  =  100     ox  bx  +  cx  —  ac  =  d 

20a7=:340     "  bx-^cx  =  d-\-ac 

(b-\-c)x=id-\-ac 

^ ^,y       u  ^ d-\-  ac 

a?  =  17       "  X  = — . 

b  +  c 
Particular  Am.  17  unbroken,  and  3  broken. 

General  Ans.    Unbroken    ^L^—, 

6  +  c 

Putting  numbers  into  the  general  answer, 

100+12  X  20_^^ 

8+12 

The  propriety  of  this  answer  may  be  shown  as  follows  :  If 
he  had  broken  the  whole  20  (or  a)  he  must  have  paid  12  X  20 
=  240  (or  a  c)  cents  ;  but  instead  of  paying  this,  he  received 
100  (or  d)  cents.  Now  the  difference  to  him  between  paying 
240  and  receiving  100  is  evidently  340,  (or  c?  +  a  c)  cents. 
The  difference  for  each  vessel  between  paying  12  and  receiving 
8  is  20  (or  6  +  c)  cents  ;  340  divided  by  20  gives  17,  the  an- 
swer. 

The  above  is  a  good  illustration  o^ positive  and  negative  quan- 
tities, or  quantities  affected  with  the  signs  +  and  — .  The 
sign  +  is  placed  before  the  quantities,  which  he  is  to  receive, 
and  the  sign  —  before  his  losses.  We  observed  that  the  dif- 
ference between  receiving  100  and  losing  240  is  340,  that  is, 
the  difference  between  +100  and  —  240  is  340,  or  their  sum. 
Also  the  difference  between  +  d  and  —  ac'i^  d  -\-  ac.  So  the 
6* 


66  AJgebra.  X. 

difference  between  +  8  and  —  12  is  20,  or  between  -j-  h  and 
—  c  is  5  -|-  c. 

Hence  it  follows,  that  to  subtract  a  quantity  which  ifas  the  sign 
— ,  we  must  give  it  the  opposite  sign,  that  is,  it  must  be  added.' 

X.  The  learner,  by  this  time,  must  have  some  idea  of  the  use 
of  letters,  or  general  symbols,  in  algebraic  reasoning.  It  has 
been  already  observed  that,  strictly  speaking,  we  cannot  actu- 
ally perform  the  four  fundamental  operations  on  these  quanti- 
ties, as  we  do  in  arithmetic ;  yet  in  expressing  these  operations, 
it  is  frequently  necessary  to  perform  operations  so  analogous  to 
them,  that  they  may  with  propriety  be  called  by  the  same 
names.  Most  of  these  have  already  been  explained  ;  but  in 
order  to  impress  them  more  firmly  on  the  mind  of  the  learner, 
they  will  be  briefly  recapitulated,  and  some  others  explained 
which  could  not  be  introduced  before. 

JVote,  Algebraic  quantities,  which  consist  of  only  one  term, 
are  called  simple  quantities,  as  -j-  2  a,  —  Sab,  &c. ;  quantities 
which  consist  of  two  terms  are  called  binomials,  as  a-{-b,a  —  b, 
3  6  4-  2  c,  &c. ;  those  which  consist  of  three  terms  are  called 
trinomials  ;  and  in  general  those  which  consist  of  many  terms 
are  c^Wed  polynomials. 

V  Simple  Quantities. 

The  addition  of  simple  quantities  is  performed  by  writing 
them  after  each  other  with  the  sign  -j-  between  them.  To  ex- 
press that  a  is  added  to  b,  we  write  a  -{-b.  To  express  that  a, 
6,  c,  d,  and  e  are  added  together,  we  write  a-\-b-\-c-\-d-\-e. 
It  is  evidently  unimportant  which  term  is  written  first,  for 
3  +  5  -{-  8  is  the  same  as  5  -|-  3  -f  8,  or  as  8  -f  5  -j-  3.  So 
a  -\-b  -\-c  has  the  same  value  as  6  -j-  a  -j-  c. 

It  has  been  remarked  (Art.  I.)  that  x  -\-  x  -\-  oc  may  be  writ- 
ten 3  X.  This  is  multiplication  ;  and  it  arises,  as  was  observed 
in  Arithmetic,  Art.  III.,  from  the  successive  addition  of  the 
same  quantity.  3  x,  it  appears,  signifies  3  times  the  quantity 
X,  that  is,  X  multiplied  by  3.  So6-]-6-f6-|-6-{-6  may  be 
written  5  6.  In  the  same  manner,  if  x  is  to  be  repeated,  any 
number  of  times,  for  instance  as  many  times  as  there  are  units 
in  a,  we  write  a  x,  which  signifies  a  times  x,  or  x  multiplied 
by  a. 


X.  Mdiiion,  Multiplication^  ^c,  67 

N.  B.  The  learner  should  constantly  bear  in  mind  that  the 
letters,  a,  b,  c,  &c.  may  be  used  to  represent  any  known  num- 
ber ;  or  they  may  be  used  indefinitely,  and  any  number  may 
afterwards  be  substituted  in  their  place. 

Again,  ab  -\-ab  -\'  ab  may  be  written  Sab,  that  is,  3  times 
the  product  a  b ;  also  c  times  the  product  a  b  may  be  written  c  a  b. 

It  may  be  remarked  that  a  times  b  is  the  same  as  b  times 
a  ;  for  «  times  1  is  «,  and  a  times  b  must  be  b  times  as  much, 
that  is,  b  times  a.  Hence  the  product  of  a  and  b  may  be  writ- 
ten either  ab  or  b  a.  In  the  same  manner  it  may  be  shown  that 
the  product  c « 5  is  the  same  els  ab  c.  Suppose  a  =r  3, 5  =  5, 
and  c  =  2,  then  «6c  =  3x5x2,  and  c  a  6  =  2  X  3  X  5.  In 
fact,  it  has  been  shown,  in  Arith.  Art.  IV.,  that  when  a  product 
is  to  consist  of  several  factors,  it  is  not  important  in  what  order 
those  factors  are  multiplied  together.  The  product  of  a,  b,  c, 
d,  e,  and/,  is  written  abcdef.  They  may  be  written  in  any 
other  order,  as  a  cdb  ef,  or  fbedca,  but  it  is  generally  more 
convenient  to  write  them  in  the  order  they  stand  in  the  al- 
phabet. 

Let  it  be  required  to  multiply  Sab  hy  2  c  d.  The  product  is 
6abcd;  for  d  times  Sab'isSabd,  but c d  times  Sabisc  times 
as  much,  or  S  abc  d,  and  2  c  d  times  Sab  must  be  twice  as 
much  as  the  latter,  that  is,  6  ab  cd. 

Hence,  the  product  of  any  two  or  more  simple  quantities  must 
consist  of  all  the  letters  of  each  quantity,  and  the  product  of  the.  coef- 
ficients of  the  quantities. 

N.  B.  Though  the  product  of  literal  quantities  is  expressed 
by  writing  them  together  without  the  sign  of  multiplication, 
the  same  cannot  be  done  with  figures,  because  their  value  de- 
pends upon  the  place  in  which  they  stand.  Sab  multiplied  by 
2cd,  for  instance,  cannot  be  written  S2  abed.  If  it  is  requir- 
ed to  express  the  multiplication  of  the  figures  as  well  as  of  the 
letters,  they  must  be  written  3  ab  2dc,orSx2  abed,  or  S.2  a  b 
c  d.  That  is,  the  figures  must  either  be  separated  by  the  let- 
ters or  by  the  sign  of  multiplication. 

Examples  in  Multiplication. 


1.  Multiply 

Sab 

2, 

5bcd 

3. 

9egh 

4. 

ISac 

5. 

25ahc 

by 

Acdf 

Ans.  12  abcdf 

by 

abc. 

Ans.  b  abbccd. 

by 

8. 

. 

by 

1  aacd. 

by 

ISabbd. 

/ 

68  Algebra.  X. 

6.  Multiply    138  by  6  a  erf. 

7.  26a?  by  11  abx, 

8.  ^^(tyy         by  \2xxy. 

It  frequently  happens,  as  in  some  of  the  above  examples,  that 
a  quantity  is  multiplied  several  times  by  itself,  or  enters  several 
times  as  a  factor  into  a  product ;  as  3  a  a  a  5  6,  into  which  a  en- 
ters three  times  and  h  twice  as  a  factor.  In  cases  like  this  the 
expression  may  be  very  much  abridged  by  writing  it  thus,  3  a^ 
h^.  That  is,  by  placing  a  figure  a  little  above  the  letter,  and 
a  little  to  the  right  of  it,  to  show  how  many  times  that  letter  is 
a  factor  in  the  product.  The  figure  3  over  the  a  shows,  that  a 
enters  three  times  as  a  factor  ;  and  the  2  over  the  6,  that  h  en- 
ters twice  as  a  factor,  and  the  expression  is  to  be  understood 
the  same  as  3  a  a  a  6  6.  The  figure  written  over  the  letter  in 
this  manner  is  called  the  index  or  exponent  of  that  letter.  The 
exponent  affects  no  letter  except  the  one  over  which  it  is 
written. 

Care  must  be  taken  not  to  confound  exponents  with  coeffi- 
cients. The  quantities  3  a  and  a^  have  very  different  values. 
Suppose  a  =  4,  then  3  a  ir=  12 ;  whereas  a^  =  4  X  4  X  4  =  64. 
In  the  product  3«^  h^  suppose  05  =  4  and  6  =  5,  then 
3  «' 6' =  3X4X4X4X5X5  =  4800. 
The  expression  a^  is  called  the  second  power  of  a,  a^  is  called 
the  third  power,  a*  the  fourth  power,  &c.  To  preserve  a  uni- 
formity, a,  without  an  exponent,  is  considered  the  same  as  a\ 
which  is  called  the  first  power  of  a.* 

Figures  as  well  as  letters  may  have  exponents. 
The  first  power  of  3  is  written 
3'  =  3 
the  second  power  3^^  =  3  X  3  =  9 

the  third  power  3^  =  3x3x3  =  27 

the  fourth  power  3*  =  3x3x3x3  =  81 

the  fifth  power  3*  =  3x3x3x3x3  =  243. 

The  multiplication  of  quantities  in  which  some  of  the  factors 
are  above  the  first  power,  is  performed  in  the  same  manner  as 
in  other  cases,  by  writing  the  letters  of  both  quantities  together, 

*  In  most  treatises  on  algebra  a2  is  called  the  square  of  a,  and  a3  the  cube 
of  a.  The  terms  square  and  cube  were  borrowed  from  geometry,  but  as  they 
are  not  only  inappropriate,  but  convey  ideas  very  foreign  to  the  present  sub- 
ject, it  has  been  thought  best  to  discard  them  entirely. 


X.  Addition f  Multiplication,  ^<j-  69 

taking  care  to  give  them  their  proper  exponents.  2  a  m*  X 
3  c'  (f  is  the  same  as  2  a  mm  X  Sccdd,  which  gives 
^ammccdd  =.  ^  am^  c^  d^, 
a^  multiplied  by  a^  gives  a^ a^  ;  but  a^  ^^aaa  and  a^  =.aa  -, 
hence  a^  a^  =i  aaaaa^=^  a!'.  In  all  cases  the  product  con- 
sists of  all  the  factors  of  the  multiplicand  and  multiplier.  In 
the  last  example  a  is  three  times  a  factor  in  the  one  quantity, 
and  twice  in  the  other  ;  hence  it  will  be  five  times  a  factor  in 
the  product.  The  exponents  show  how  many  times  a  letter  is 
a  factor  in  any  quantity  ;  hence  if  any  letter  is  contained  as  a  fac- 
tor one  or  more  times  in  both  multiplier  and  multiplicand,  the  expo- 
nents being  added  together  ivill  give  the  exponent  of  that  letter  in 
the  product. 

Ans.  a^  b\ 


ax  a  = 

a'  Xa'  =  a'+'  =  a\ 

a^  X  a^  =  « 

a'Xa'  =  a'^'  = 

:  a\  &c. 

9. 

Multiply 

a'b'               by 

ab\ 

10. 

a¥c             by 

aHcK 

11. 

ea'cd"        by 

ab'c\ 

12. 

a'  e              by 

d'b'c. 

13. 

1  a^x^y        by 

5a^bcx'y\ 

14. 

17  b^  d^  e       by 

Abbe  dee. 

15. 

2Sa'x^         by 

2aabxx. 

16. 

IS  aayy     by 

Qa^yyx. 

It  has  already  been  remarked  that  the  addition  of  two  or 
more  quantities  is  performed  by  writing  the  quantities  after 
each  other  with  the  sign  +  between  them.  The  sum  of  3  «  b^ 
2acd,  5  a^b,  4ab,  and  3a^6,  is3a6  -}'2acd-\-5a^b-\-'4ab 
+  3  a^  6.  But  a  reduction  may  be  made  in  this  expression,  for 
3  ab  -{-  4  ab  is  the  same  as  7  a  6  ;  and  5  a^  6  -j-  3  a*  6  is  the  same 
as  8a^b;  hence  the  expression  becomes 

7«6-f  2  ac<7  + 8  ft' 6. 

Reductions  of  this  kind  may  always  be  made  when  two  or 
more  of  the  terms  are  similar.  When  two  or  more  terms  are 
composed  of  the  same  letters,  the  letters  being  severally  of  the 
same  powers,  they  are  said  to  be  similar.  The  numerical  co- 
efficients are  not  regarded.  The  quantities  4ab  and  Sab  are 
similar,  and  so  are  5  a^  b  and  Sa^b  ;  but  4ab  and  5a^b  are  not 
similar  quantities,  and  cannot  be  united. 

The  subtraction  of  algebraic  quantities  is  performed  by 
writing  those,  which  are  to  be  subtracted,  after  those  from 
which  they  are  to  be  taken,  with  the  sign  —  between  them. 


70  Algebra,  XL 

If  5  is  to  be  subtracted  from  a  it  is  written  a —  h.  5  a  6^  to  be 
subtracted  from  8  a  b^,  is  written  8ab^  —  5ab^,  This  last  ex- 
pression may  be  reduced  to  3  a 6'.  In  all  cases  when -the 
quantities  are  similar,  the  subtraction  may  be  performed  imme- 
diately upon  the  coefficients. 

Compound  Quantities. 

XL  The  addition  and  subtraction  of  simple  quantities,  pro- 
duce quantities  consisting  of  two  or  more  terms  which  are 
called  compound  quantities.  2a  -}-  c  d  —  36  is  a  compound 
quantity. 

Addition  of  Compound  Quantities. 

The  addition  of  two  or  more  compound  quantities,  when  all 
the  terms  are  affected  with  the  sign  -f-  will  evidently  be  the 
same,  as  if  it  were  required  to  add  together  all  the  simple  quan- 
tities of  which  they  are  composed  ;  that  is,  they  must  be  writ- 
ten one  after  the  other  with  the  sign  -j-  before  all  the  terms  ex- 
cept the  first.  The  sum  of  the  quantities  3  a  -}-  2  c  and  b  -{-2d 
is3a-^2c-{-b  -\-2d. 

If  the  quantities  3  a  6  -\-  5  d  and  b  —  c  be  added,  in  which 
some  of  the  terms  have  the  sign  — ,  the  sum  will  be  3  a  6  +  5  rf 
-f-  6  ■*—  c  ;  for  6  —  c  is  less  than  b,  therefore,  if  b  be  added  the 
sum  will  be  too  large  by  the  quantity  c.  Hence  c  must  be  sub- 
tracted from  the  result. 

This  may  be  illustrated  by  figures.     Add  together  17-j-  10 
and  20  —  6.     Now  20  —  6  is  14 
and  17  +  10  4-  20  —  6  is  equal  to  17  +  10  -f-  14. 

From  the  above  observations  we  derive  the  following  rule  for 
the  addition  of  compound  quantities. 

Write  the  quantities  after  each  other  without  changing  their  signs, 
observing  that  terms  which  have  no  sign  before  them  are  understood 
to  have  the  sign  -f-. 

A  sign  affects  no  term  except  the  one  immediately  before 
which  it  is  placed  ;  hence  it  is  unimportant  in  what  order  the 
terms  are  written,  for  14  —  5  -j-  2  has  the  same  value  as  14  -f- 
2  —  5  or  as  —  5-}-2  +  14.  Those  which  have  the  sign  -j- 
are  to  be  added  together,  and  those  which  have  the  sign  —  are 
to  be  subtracted  from  their  sum.     If  the  first  term  has  the  sign 


XI,  J       Addition  of  Compound  Quantities.  71 

4-,  the  sign  may  be  omitted  before  this  term,  but  the  sign  — 
must  always  be  expressed.  Great  care  is  requisite  in  the  use 
of  the  signs,  for  an  error  in  the  sign  makes  an  error  in  the  re- 
sult of  twice  the  quantity  before  which  it  is  written. 

Add  together    3a -\-2bc^  —  Sc' 
and  5a  —  2bc'-\-2c* 

and  7a6  +  46c'— 8c* 

and  ^a  -\-Sc'-^2bc\ 

Th^um  is 

3a  +  2bc'-^Sc'  +  5a—'Shc'  +  2c'-}-7ab 
j^  4h  c'  ^8  c'  —  a  +  3  c'  '-2b  c\ 
But  this  expression  may  be  reduced. 

3  a  -{-  5  a  —  a  =-8  a  —  a=:7  a, 
and 

2b  c'  --  3b  c'  -{-  4b  c^  -^2b  c'  =  6b  c'  --  5b  c'  =zb  c% 
and 
—  3  c*  -f  2  c*  —  8  c*  +  3  c*  ==  —  1 1  c*  -h  5  c*  =  —  6  c* ; 

hence  the  above  quantity  becomes 

1a-{-bc^-^7  ab^6c\ 

To  reduce  an  algebraic  expression  to  the  least  number  of 
terms,  collect  together  all  the  similar  terms  affected  with  the  sign  -\- 
and  also  those  affected  with  the  sign  — ,  and  add  the  coefficients  of 
each  separately ;  take  the  difference  of  the  two  sum^  and  put  it  into 
the  general  residt,  giving  it  the  sign  of  the  larger  quantity. 

Examples  in  Addition. 

1.  Add  together  the  following  quantities. 

6  ab  —  2a^m  *- 

and  3  ab  —  5  am -^2  am. 

2.  Add  together  the  following  quantities. 

13  a  n^  —  6  m  -{-x^, 
and  7bm  —  3x^  —  8y, 

and  4  a  n*  +  5  a  a;^  —  4y. 


72  Algebra.  XII. 

3.  Add  together  the  following  quantities. 

1  mab  —  16  —  4Smy, 
and  19  €icb — IS  a mb -\- 37 ma y -{- 48, 

and  14  my — 19may-\-nb  —  nx, 

and  4nx  —  Sbn-\-22amy  —  nb, 

4.  Add  together  the  following  quantities. 

a?y  —  ax  —  ay  -{-  axy, 
and  — '2>xy  —  2ay  -\-Sax  -\-\^, 

and  I8arx  —  lS-\-lSaxy  —  am, 

and  — \baxy — 13a»i -f- 43 -{- 18ara?, 

and  arx  —  IS -\- ay — 2axy-\-Sam. 

6.  Add  together  the  following  quantities. 

ISao:  — 26a?  — 7, 
and  Ibbx — lib xy  -^  l^, 

and  41  acd  —  x, 

and  37 — bx  —  2a-\-4Sbyx, 

and  acd-^-byx-^lSa, 

Subtraction  of  Compound  Quantities. 

XII.  The  subtraction  of  simple  quantities,  as  has  already 
been  observed,  is  performed  by  giving  the  sign  —  to  the  quan- 
tity to  be  subtracted,  and  writing  it  before  or  after  the  quau- 
trty,  from  which  it  is  to  be  taken.  If  it  is  required  to  subtract 
c  -\-  d  from  a  -f  ^  i^  is  plain  that  the  result  will  hea-\-b  —  c 
—  d,  for  the  compound  quantity  c-{-  d'ls  made  up  of  the  sim- 
ple quantities  c  and  d,  which  being  subtracted  separately  would 
give  the  above  result. 

From  22  subtract  13  —  7. 

13  —  7  =  6. 

and  22  —  6  =  16. 

The  result  then  must  be  16.  But  to  perform  the  operation 
on  the  numbers  as  they  stand,  first  subtract  13,  which  gives 
22  —  13  =  9.  This  is  too  small  by  7  because  the  number  13 
is  larger  by  7  than  the  number  to  be  subtracted,  therefore  in 
order  to  obtain  a  correct  result  the  7  must  be  added ;  thus  22 
— 13+7  =  16,  as  required. 


XII.  Subtraction  of  Compound  Quantities,  73 

From  a  subtract  b  —  c. 

First  subtract  b,  which  gives  a  —  b. 

This  quantity  is  too  small  by  c  because  b  is  larger  than  b — c 
by  the  quantity  c.  Hence  to  obtain  a  correct  result  c  must  be 
added,  thus  a  —  b-\-  c. 

This  reasoning  will  apply  to  all  cases,  for  the  terms  affected 
with  the  sign  —  in  the  quantity  to  be  subtracted  diminish  that 
quantity  ;  hence  if  all  the  terms  affected  with  -f-  be  subtract- 
ed, the  result  will  be  too  small  by  the  quantities  affected  with 
— ,  these  quantities  must  therefore  be  added.  The  reductions 
may  be  made  in  the  result,  in  the  same  manner  as  in  addition. 
Hence  the  general 

Rule.  Change  all  the  signs  in  the  number  to  be  subtracted,  the 
signs  -\-  to  — ,  and  the  signs  —  to  -\-,  and  then  proceed  as  in  ad- 
dition. 

Examples  in  Subtraction. 

1.  From        a*a?  +  35y  — 5ac'— 16 
Subtract  3a*a:  + 5y  — 2ac'-— 22 

Operation. 

a^ X  -\-Sby  —  5ac^  —  16 
^Sa'x'—by  +  2ac^  +  22 


— 

-2a'a:-|-26y  — 3ac'-f  6 

From 
Subtract 

Sbaf-^lax'^lS 
lSbc  —  Sax'—8. 

Ans.  Sbx*  —  lSbc 

From 
Subtract 

17a'y  +  13a/  — a  — 3 
2a'y— &  — lla  +  5.   ' 

From 
Subtract 

42  axy  —  4 ax 
llax  —  2axy  —  5. 

From 
Subtract 

143— 17y 

33  +  4y  — I6a&. 

7 

4a^='  +  21. 


74  Algebra.  XIII. 

6.  From  a-\'^ahc  —  1 
Subtract  1  ■^-^ahc  —  a, 

7.  From  3a5^  +  2a5-— 7z 
Subtract  2  ah  —  Iz  —  2ahz, 

Mvliiplicaiion  of  Compound  Quantities, 

XIII.  Multiplication  of  compound  quantities  is  sometimes 
expressed  without  being  performed.  To  express  that  a  -[-5  is 
to  be  multiplied  by  c —  d,  it  may  be  written  a  +?  X  c— 5 
with  a  vinculum  over  each  quantity,  and  the  sign  of  multiplica- 
tion between  them ;  or  they  may  be  each  enclosed  in  a  paren- 
thesis and  written  together,  with  or  without  the  sign  of  multi- 
plication ;  thus  (a  +  6)  X  {c  —  d)  or  {a  -j-  h)  (c  —  d).  In  the 
expression  a  -\-h  (c — d)^h  only  is  to  be  multiplied  by  c  —  d. 

Multiply  a-\-hhy  c. 

It  is  evident  that  the  whole  product  must  consist  of  the  pro- 
duct of  each  of  the  parts  by  c. 

a +6  20-^-4    =24 

c  3  3 


ac+hc  60+12  =  72 

Examples, 

3ah  +  2cdhy  ef, 

Ans,  Sabef  +  2cdef, 
5ac-|-6c  +  3c(Zby2e. 

Ans.  10  ace-\- 2b  ce-^6cde. 
Qa'h-^b'c'hySab'. 
bc'd'-^52a'h'  +  l3b'c*d 

laH'c, 
2ahd  +  '^abx^  +  aboc^' 
Zab(x^, 
ax'  +  ^abx^    by     13a  6*^' 

When  some  of  the  terms  of  the  multiplicand  have  the  sign 
•  they  must  retain  the  same  sign  in  the  product. 


1.  Multiply 

2.  Multiply 

3.  Multiply 

4.  Multiply 

by 

5.  Multiply 

by 

6.  Multiply 

^III.  Multiplication  of  Compound  Quantities. 

7.  8.  Multiply  a  —  6  by  c,  also  23  —  5  by  4. 

a--5  23  —  5    =18 

c  4  4 


7e5 


a  c —  be. 


92— -20  =  72. 


Since  the  quantity  a  —  6  is  smaller  than  a  by  the  quantity 
b,  the  product  a  c  will  be  too  large  by  the  quantity  b  c.  This 
quantity  must  therefore  be  subtracted  from  a  c. 


9.  Mult] 

ply 

3ab'  —  c 

by 

2d. 

10.         " 

2ad-{'bd  —  Sc 

by 

dab. 

11.         « 

Sbcd^ef'-2ac 

by 

d  ac. 

12.         « 

2a'be^5a'-\-b' 

by 

4a't 

13.         « 

I7ac</— 1  ^5a^x  — 
by                  a^  c  dr 

-ab' 

X 

When  both  multiplicand  and  multiplier  consist  of  several 
terms,  each  term  of  the  multiplicand  must  be  multiplied  by 
each  term  of  the  multiplier. 


14.  Multiply 


12  +  5 
12  -J-  5  =  17 
7  +4  =  11 


by    7+4. 


84+35 
+  48  +  20 

84  +  35+48+20  =  187 

15.  Multiply    a +  6 

a^b 
.  c  +  d 


by     c  -\-d. 


ac-\-bc-\-ad-{-bd. 

It  is  evident  that  if  a  +  &  be  taken  c  times  and  then  d  times, 
and  the  products  added  together,  the  result  will  be  c  +  <?  times 
a +  6. 


76  Algebra,  XIII. 

16.  Multiply     ax  —  Say  +  xy    by    day-^-ax. 
ax— -Say  -\-  xy 
Say  -\-  a  X 

S(^xy'—9a^y'  +  3axy* 
a^ x^  —  3a^ xy  -\'ax^y 


a*  X*  —  9  a^  'f  -\-  S  a  y  x^  -\-  a  x^  y. 

In  adding  these  two  products,  the  quantity  3  a^  a;  y  occurs 
twice,  with  different  signs ;  they  therefore  destroy  each  other 
and  do  not  appear  in  the  result. 

17.  Multiply       bad  +  Sacd —  5a^c 

by  2a'c-j-2aflf/ 

18.  Multiply     ISaSy  — 2a6  2/^  +  3cy' 

by  5c/  +  7rt6/+3. 

19.  Multiply     1 1  a  c'  -f.  3  a'  c  —  4  a' 

by  2  a*  c  4-  a  c' 

20.  Multiply     a.2  _  2  a  c  +  c'         by         a  +  c. 

21.  «  Sa'  —  Sh'  by         2  a' +  2  6^ 

22.  «  2b  +  2c  by         2a-^3^. 

3J+2C 
2a--36 


6ab  +  4ac 
—  96'  — 6Jc 


6a6-f-4ac  — 96'  — 66c. 

If  3  6  +  2  c  be  multiplied  by  2  a  only,  the  product  will  be 
too  large  by  3  6  times  3  6  +  ^^5  hence  this  quantity  must 
be  multiplied  by  3  6,  and  the  product  subtracted  from  6  a  6  -j- 
4ac. 

This  result  may  be  proved  by  multiplying  the  multiplier 
by  the  multiplicand,  for  the  product  must  be  the  same  in  both 
cases. 

23.  Multiply     2ad'{-Sbc^2     by     4a 6  — 2c. 


(  UNiV 


n B  R  A  RY 

^  Of  THE 


UNIVERSITY 


XIII.  Multiplication  of  Compound  Quantities,  11 

24.  Multiply     6«'J  +  2a5'    by     2a'b'—b''^l. 

25.  «  19  —  5  by  9  —  4. 

19  —  6  =14 

9  —  4  =    5 


171  —  45  -  70. 

—  76  +  20 

171  __  45  —  76  +  20  =  191  —  121  =  70. 

26.  Multiply    a  —  h        by        c  —  d. 
a  —  b 
c — d 


ac  —  be 
— ^fl<?  +  bd 


ac  —  be  —  ad  -\-bd. 


This  operation  is  sufficiently  manifest  in  the  figures.  In  the 
letters,  I  first  multiply  a  —  bhy  c,  which  gives  a  c  —  be;  but 
the  multiplier  is  not  so  large  as  c  by  the  quantity  d,  therefore 
the  product  ae  —  5  c  is  too  large  by  d  times  a  —  b;  this  then 
must  be  multiplied  by  d  and  the  product  subtracted,  a  —  b 
multiplied  by  d  gives  ad — bd;  and  this  subtracted  from 
ae  —  be  gives  ae  —  be  —  ad  -{-  bd.  Hence  it  appears  that 
if  two  terms  having  the  sign  —  be  multiplied  together,  the  pro- 
duct must  have  the  sign  +. 

From  the  preceding  examples  and  observations,  we  de- 
rive the  following  general  rule  for  multiplying  compound  quan- 
tities. 

1.  Multiply  all  the  terms  of  the  multiplicand  by  each  term  of  the 
multiplier f  observijig  the  same  rules  for  the  coefficients  and  letters  as 
in  simple  quantities, 

2.  With  respect  to  the  signs  observe, 

1st,  That  if  both  the  terms  which  are  multiplied  together ,  have 
the  sign  -{-,  the  sign  of  the  product  must  be  -}-. 

2d,  If  one  term  be  affected  with  -f-,  and  the  other  with  — ,  tlie 
product  must  have  the  sign  — . 
7* 


78  Akf^a-  XIII. 

3d,  If  both  term  he  affected  with  the  sign  — ,  the  proiuct  must 
have  the  sign  +. 

Or  in  more  general  terms,  If  both  tei-ms  have  the  same  sign, 
whether  -|-  or  — ,  the  product  must  have  the,  sign  +,  and  if  they 
have  different  signs,  the  product  must  have  the  sign  — . 

27.  Multiply     3a*^>— 2ac  +  5 
by  1  ab  —  2a  c  —  1. 


2laH^'-Ua'bc  +35«6 

^2a'b  +  2aC'—5. 

Product 

2la='62  — 14a'6c  +  35ai  — 6fl=^6c-i-4a'c'— 8«c— 3a'i— -5. 


28. 

Multiply 

7m  +  5  n 

by 

4wi  — 3w. 

29. 

it 

a^ -\-  ay  —  / 

by 

a  —  y. 

30. 

ii 

ji^^ncr-f  0?' 

by 

n--x. 

31. 

(C 

a^^ab-^-h' 

by 

a'^ab  +  b\ 

32. 

(( 

2x'  —  Sxy  +  4y' 

by                 5  a? 

—  ( 

5/ry  — 

•2y=. 

33. 

ii 

3a«c--5ac'-|-2 

c' 

by                 2  a' 

c  — 

-4a*c^ 

'  — 7ac'. 

34. 

u 

2a^  —  a'x  +  2 

by 

3a- 

-a:  — 3. 

35. 

ii 

7a^6  +  2  6^  — 1 

by 

3a^- 

-26^—1. 

It  is  generally  much  easier  to  trace  the  effect  produced  by 
each  of  several  quantities  in  forming  the  result,  when  the  ope- 
rations are  performed  upon  letters,  than  when  performed  upon 
figures.  The  following  are  remarkable  instances  of  this.  They 
ought  to  be  remembered  by  the  learner,  as  frequent  use  is 
made  of  them  in  all  analytical  operations. 

Let  a  and  b  represent  any  two  numbers ;  a-^-b  will  be  their 
sum  and  a — b  their  difference. 


XIII.  Multiplication  of  Compound  Quantities.  79 

Multiply    a  +  b  by  a -^6. 

a  +  b 
a  —  b 


—  ab-^b' 

a'^b\ 

That  is,  if  the  sum  and  the  difference  of  two  numbers  be  multipli- 
ed together  J  the  product  will  be  the  difference  of  the  second  povjers 
of  these  two  nunwers. 

Particular  Example. 

Let  a  =  12and5=T. 

a  +  5  =  l9,  andfl  —  6  =  5, 
a*  =  144,  6'  =  49.  ^ 

{a  +  b)x  (a  — 6)=:19X  5  =  95, 
and 

Multiply   a+b   by   a  -j-  6. 
a+b 
a  +  b 


a^  +  ab 

ab  +  b' 

a^^2ab  +  b\ 

That  is,  the  product  of  the  sum  of  two  numbers,  by  itself,  or  the 
second  power  of  the  sum  of  two  numbers,  is  equal  to  the  sum  of  the 
second  powers  of  the  two  numbers,  added  to  twice  the  product  of  the 
two  numbers. 

Multiply     a — b  by  a-^b. 

The  answer  is  a'  —  2ab  +  b*, which  is  the  same  as  the  last, 
except  the  sign  before  2ab, 

Multiply  a^+2ab  +V  hy  a+  b,  that  is,  find  the  third 
power  of  a +5. 

Ans,  a'  +  3a*5  +  3 a 6'  +  6'. 


80  Algebra.  XIV.. 

This  is  expressed  in  words  thus  :  the  third  power  of  the  first, 
plus  three  times  the  second  power  of  the  first  into  the  second,  plus 
three  times  the  first  into  the  second  power  of  the  second,  plus  thejhird 
power  of  the  second. 

Multiply  a*  —  2  a  6  -h  &*     by     a  —  5. 

Ans.  a'  — 3a*6  +  3a6'  — 6'. 

Which  is  the  same  as  the  last,  except  the  signs  before  the  se- 
cond and  last  terms. 

Instances  of  the  use  of  the  above  formulas  will  frequently 
occur  in  this  treatise. 

Dividon  of  Algebraic  Quantities. 

XIV.  The  division  of  algebraic  quantities  will  be  easily  per- 
formed, if  we  bear  in  mind  that  it  is  the  reverse  of  multiplica- 
tion, and  that  the  divisor  and  quotient  multiplied  together  must 
reproduce  the  dividend. 

The  quotient  of  a  5  divided  by  a  is  b,  for  a  and  b  multiplied 
together  produce  ab.  So  ab  divided  by  b  gives  a  for  a  quo- 
tient, for  the  same  reason. 

If  6  a  6  c  be  divided  by  2  a,  the  quotient  is  3  fe  c. 
If  by  2  b,  the  quotient  is  3  a  c. 

If  by  2  c,  the  quotient  is3  ab. 

If  by  3  b,  the  quotient  is  2  a  c. 

If  by  S  ab,  the  quotient  is  2 c. 

If  by  6  a  the  quotient  is  6  c. 

For  in  all  these  instances  the  quotient  multiplied  by  the  di- 
visor, produces  the  dividend  6  abc. 

Examples. 

1.  How  many  times  is  2  a  contained  in  6  a  6  c? 
Ans.  3  6c  times,  because  36c  times  2  a  is  6  a  6  c. 

2.  If  6  a  6  c  be  divided  into  2  a  parts,  what  is  one  of  the 
parts  .'' 

Ans.  2b  c;  because  2  a  times  3  6  c  is  6  a  6  c. 


XIV.  Division  of  Algebraic  ^antities,  8 1 

Hence  we  derive  the  following 

Rule.  Divide  the  coefficient  of  the  dividend  by  the  coefficient  of 
the  divisor,  and  strike  out  the  letters  of  the  divisor  from  the  divi- 
dend, 

3.  Divide  16  ab  c 

4.  "  12  abc 

5.  *<  20  abc 

6.  "  18  abed 

7.  "  22  abc 

8.  "  Had 

9.  "  4  a' 

Observe  that  4  a^  is  the  same  as  4  a  a  a  and  a^  is  the  same 
as  a  a  ;  4  a  aa  divided  by  a  a  gives  4  a  for  the  quotient. 

It  was  observed  in  multiplication,  that  when  the  same  letter 
enters  into  both  multiplier  and  multiplicand,  the  multiplication 
is  performed  by  adding  the  exponents,  thus  a^  multiplied  by  a* 
is  a^^  =z  «^  In  similar  cases,  division  is  performed  by  subtract- 
ing the  exponent  of  the  divisor  from  that  of  the  dividend,     a^  divid- 


by 

4. 

by 

3  a. 

by 

10  be. 

by 

6  ad. 

by 

ab. 

by 

ad. 

by 

a'. 

edby 

a'isa'-' 

^=z«^ 

10. 

Divide 

6a'b'c 

by 

Sab\ 
Ans,  2 abc. 

11. 

u 

S5b'd^ 

by 

bd. 

12. 

li 

16  «V 

by 

4a'c\ 

13. 

<( 

18  a;*/ 

by 

ey*. 

14. 

u 

48a*x'm 

by 

I6a'j:m 

15. 

(C 

72ar'm' 

by 

12  a  r\ 

16. 

u 

60pV 

by 

60. 

17. 

(( 

730./ 

by 

ap\ 

18. 

(( 

120  a  r't' 

by 

rf. 

The  division  of  some  compound  quantities  is  as  easy  as  that 
of  simple  quantities. 

If  a  +  ^  +  c  be  multiplied  by  d  the  product  is 
d{a-\'b-^c)  ox  a  d  -\-b  d  -\'  c  d. 

Therefore  ifa<?  +  6(?  +  c(?be  divided  by  </,  the  quotient  is 
a-\-b^c. 


91  Algebra,  XV. 

l{ad+hd-\'Cdhe  divided  by  a  +  5  +  c,  the  quotient  is  d. 

When  a  compound  quantity  is  to  be  divided,  let  the  quan- 
tity, if  possible,  be  so  arranged  that  the  divisor  may  appear  as 
one  of  the  factors,  and  then  that  factor  being  struck  out,  the 
other  factor  will  be  the  quotient. 

19.  Divide     12  a'  6  —  9  a  c     by  3  a. 

12a'6  — 9ac  =  3a(4a6--3c) 

Ans,  4ab  —  3c. 

Observe  that  a  is  a  factor  of  both  terms,  and  also  3.  Hence 
the  quantity  12  a^b  —  9  a  c,  can  be  resolved  into  factors  ;  thus 
3  (4  a'  6  —  3  fl  c),  or  a  (12  a  6  —  9  c),  or  3  a  (4  a  6  —  3  c).  In 
the  last  form  the  divisor  3  a  appears  as  one  factor,  and  the  other 
factor  4  ah  —  3 c  is  the  quotient. 

JSTote.  Any  simple  quantity,  which  is  a  factor  of  all  the  terms 
of  any  compound  quantity,  is  a  factor  of  the  whole  quantity  ; 
and  that  factor  being  taken  out  of  all  the  terms,  the  terms  as 
they  then  stand,  taken  together,  will  form  the  other  factor. 

20.  Divide      8aH^—l6a'b^c     by     2a6  — 4a'c. 

8a'5'— 16a'6'c=:4a6*(2a6— 4a'c.) 

Ans.  4  a  6*. 

21.  Divide      Sabc-^W  ab' d-\- 9  a'bd'     by     Sab, 

22.  Divide     I5a^bc--30a'c^  +  26a'cd 

by  5a'c. 

23.  Divide     36  a''  6*  c  —  28  a"  6V  +  40  a'  b'  ^ 

by  9a«— 7a*6»c+10a^&*c*. 

24.  Divide     42  a' —  84  a'°  6' c     by     1— 2a'6'c. 

Algebraic  Fractions, 

XV.  When  the  dividend  does  not  contain  the  same  letters 
as  the  divisor,  or  but  part  of  those  of  the  divisor,  the  division 
cannot  be  performed  in  this  way.  It  can  then  only  be  express- 
ed. The  usual  way  of  expressing  division,  as  has  already 
been  explained,  is  by  writing  the  divisor  under  the  dividend  in 

the  form  of  a  fraction.     Thus  a  divided  by  b  is  expressed   -. 

6 


XV.  Algebraic  Fractions.  83 

This  gives  rise  to  fractions  in  the  same  manner  as  in  €irithme- 
tic.  It  was  shown  in  arithmetic,  that  a  fraction  properly  ex- 
presses a  quotient.  Algebraic  fractions  are  subject  to  precise- 
ly the  same  rules  as  fractions  in  arithmetic.  Many  of  the  ope- 
rations are  more  easily  performed  on  algebraic  fractions. 

In  these,  as  in  arithmetic,  it  must  be  kept  in  mind,  that  the 
denominator  shows  into  how  many  parts  a  unit  is  divided  ;  and 
the  numerator  shows  how  many  of  those  parts  are  used  ;  or  the 
denominator  shows  into  how  many  parts  the  numerator  is  di- 
vided. 

I  shall  here  briefly  recapitulate  the  rules  for  the  operations 
on  fractions,  referring  the  learner  to  the  Arithmetic  for  a  more 
full  developement  oftheir  principles. 

2  times  A  =  A. 
11       11 

2time8?  =  ?f. 


..        a      ac 
c  times  _  =  _-. 

b        b 
I  of  7  is  V  ;  for  j  of  7  is  J,  and  |  is  3  times  as  much,     f  of 
a  is  ^  * ;  for  4  of  a  is  ^,  and  f  is  2  times  as  much.     The  ^L  part 
of  c  is^ ;  for  y  of  c  is  -^,  and  j  is  a  times  as  much. 

Hence,  to  multiply  a  fraction  by  a  whole  number,  or  a  whole 
number  by  a  fraction,  multiply  the  numerator  of  the  fraction  and  the 
whele  number  together,  and  divide  by  the  denominator, 

AHth, .Articles  XV.  &  XVI. 

Examples, 

1.  Multiply  f+i  by  2.         Am.  ?±+ll. 

c  c 

2.  Multiply  la+2^        y^^  ^^ 


ac 


Am,  ^^^d  +  ^^'d 
ac 


64 


3.  Multiply 

4.  Multiply 

5.  Multiply 

6.  Multiply 


Algebra. 

Sbc — 2fl 
6a — 13  c 


2ab  —  be 

3ab 
5  a  c  —  2  m* 

13ac~ 

16  a x^  —  Sbx 


XVI. 


by 

by 
by 
by 


4  b'. 
12b'c  —  8ab' 


5a- 

--13c 

5ac 

+  3c^ 

5a5 

— .3w. 

2m- 

—  3a; 

2  a  +  7  a;' 


Division  of  Fractions. 


XVI.     1.  Divide 

4a 

7 

by 

2,     or  find  J  of 

4a 

7 

An..  ^«. 
7 

2.  Divide 

06 

by 

a,      or  find  _    of 

c 

a 

ab 

Ans,      t 

c 

c 

3.  Divide 

6a*6 

by 

3  a,     or  find of 

erf 

3a 

cd 

Ans.      ^^\    ' 
cd 

4.  Divide 

f  "y 

2,     < 

or  find  J  of  - . 

0 

This  cannot  be  done  like  the  others,  but  it  may  be  done  by 
multiplying  the  denominator  as  in  Arith.  Art.  XVII.  For  the 
fraction  ^  denotes,  that  one  is  divided  into  as  many  equal  parts 
as  there  are  units  in  b,  and  that  as  many  of  these  parts  are  used 
as  there  are  units  in  a  ;  or  that  a  is  divided  into  as  many  equal 
parts  as  there  are  units  in  b ;  hence  if  it  be  divided  into  twice 
as  many  parts,  the  parts  will  be  only  one  half  as  large,  and  the 
fraction  will  have  only  one  half  the  value. 


XVI.  Division  of  Fractions.  85 

Hence  -  divided  by  2,  is  iL. 

So        -  divided  by  c?,  is  — 
c  cd 

5.  Divide     1^    by     ^d.        Am.     —. 

2cd      ^  8cd* 

Hence,  to  divide  a  fraction  by  a  whole  number^  divide  the  nu- 
merator ;  or  when  that  cannot  he  done^  multiply  the  denominator  by 
the  divisor. 

6.  Divide        ^  by  3  a. 

d 

7.  Divide         11^  by  7aL 

cd 

8.  Divide        ^^l^i  by  2a'c. 

3dm  ^ 

9.  Divide        ?^  by  h\ 

bac 

10.  Divide       Hi  by  3.  . 

a 

11.  Divide       i^  by  5. 

6  d 


12 


.  Divide       ^^^  by  2hm. 


mnr 


13.  Divide       ii^  by  3c*£?. 

14.  Divide       ^1^  by  Qa'ch. 

lac^d  ^ 


15. 


Divide       iiZli^  by  3arf. 


26c 


16.  Divide      7a^--mc     ^^  3^^ 

2ad^5b'  ^ 


17.  Divide  ^^^^^    ■       by  Thn 

ban  —  w  n* 

8 


86                                      Mgd 

frai 

XVIK 

18.  Divide        ^V^^"^'. 

by 

4o*  +  ^ni 

19.  Divide          ^^«^ 
46  +3  a: 

by 

4  6-«.^a?. 

20.  Divide              ^«--/ 

3a  — 4crf+l 

by 

Ta  +  ^crf'^^i; 

21.  What  is  1  of  ^?     J  of? 
0                  6 

S-6 

and' 

|is  twice  as  much. 

that  is,  ||. 

22.  What  is  the  %  part  of  f  ? 

.i-l 

is  -£-  and  ?   is  a 
od          0 

times  as  much,  that  is,  ^. 
od 

That  is,  4.  X   -^  ^  ^. 

Hence,  to  multiply  one  fraction  by  another,  multiply  the  nume- 
rators together  for  a  new  numerator,  and  the  denominators  together 


for  a  new  denominator. 

Jlrith,  Art,  XVII, 

23.  Multiply 

3c     ^ 

b 
2m' 

An.,^^K 
6cm 

24.  Multiply 

Sad 
46c 

by 

Sam 
ben 

25.  Multiply 

12  an  X 
ISbry 

by 

Sao? 
26y* 

26.  What  is 

2a*m 
bed 

of 

3mrf 

27.  What  is 

Ab'd 
^ax' 

of 

76m   p 
13?ia? 

28.  Multiply 

2a 
36+c 

by 

2ac— 5c 
5a6 

29.  Multiply 

2am*- 
4ac 

-3a'm  ,          bam^ 
-|-2c        "  2am  — 5c 

J^VJ.  Multiplication  and  Division  of  Fractions,  87 

30.  Multiply  2«c  +  3    by     3«6 


5bc-—2(P         dac^2ad    ' 

31.  Multiply  2a^m-\-Sm^  ^y  1^ 

We  have  seen  that  a  fraction  may  be  divided  hy  multiplying 
its  denominator,  because  the  parts  are  made  smaller  ;  on  the 
contrary,  a  fraction  may  be  multiplied  by  dividing  its  denomi- 
nator, because  the  parts  are  made  larger.  Arith.  Art.  XVIII. 
If  the  denominator  be  divided  by  2,  the  unit  is  divided  into 
only  one  half  as  many  parts  ;  consequently  the  parts  must  be 
twice  as  large  as  before.  If  the  denominator  be  divided  by  5, 
the  unit  is  divided  into  only  one  fifth  as  many  parts  ;  hence  the 
parts  must  be  five  times  as  large  as  before,  and  if  the  same  num- 
ber of  parts  be  used  as  at  first,  the  value  of  the  fraction  will  be 
five  times  as  great  and  so  on. 

32.  Multiply  ^—        by         6.       ,dns,  ^. 
\  20  4 

33.  Multiply  ^         by        6. 

If  we  divide  the  denominator  by  b,  the  fraction  becomes 

1 
— ,  in  which  a  is  divided  into  — -  part  as  many  parts  ;   hence 
c  0 

the  part$, -and  con$ejqu^ntly  the  .fraction  is  &.  times  as  large  as 
before. 

34.  Multiply  -^  by        2  c. 

6o  c 

35.  Multiply  Jl^  by         8(fd, 

36.  Multiply  ^^  by         7am^ 

42a^m^ 

37.  Multiply  ^^  by         5 ma;. 

25  m*a: 

38.  Multiply  ~  by        5. 

5a 


88  Algehra.  XVI. 

39.  Multiply  JL_  by        ab. 

8ab  ^ 

40.  Multiply  f"^  ■—       by        46. 

41.  Multiply  ^^"~t,^^ by    Aa\ 


42.  Multiply 


23  7W— 13 


by  7  m^  c. 

43.  Multiply       f  by  5. 

Dividing  the  denominator  by  5  it  becomes  f ,  or  3. 

Multiply  -J  by  b. 

b 

Dividing  the  denominator  by  b  it  becomes  —,  or  a. 

44.  Multiply       i^,    by     5  6  c?.     ^m.  ii5i  =  3  a  c. 

56  a  1 

1  h  n 

In  fact  —-  multiplied  by  6  is  ^  =  1,  and  Z.  being  a  times 
o  6  6 

as  much  as  _.,  must  give  a  product  a  times  as  large,  or  a 
6 

times  1,  which  is  a. 

Hence,  if  a  fraction  be  multiplied  by  its  denominator,  tJte  pro- 
duct  will  be  the  numerator. 

45.  Multiply  !i£?*  by  5  6df. 

^'  bbd  ^ 

46.  Multiply  ^  by  36  c. 

36c 

47.  Multiply  iif-  by  4  6m^ 

4  6  m^ 

48.  Multiply  12mY        ^y  bdn'x, 

bdr?  X 


XVII.  Reducing  Fractions  to  Lower  Terms.  89 

49.  Multiply 

50.  Multiply 

51.  Multiply 

Two  ways  have  been  shown  to  multiply  fractions,  and  two 
ways  to  divide  them. 

To  multiply  a  fraction,}^  j.  ,       ^  the  numerator 

To  divide  a  fraction,     )  ^^  ^      X  the  denominator. 


13a6  — i/i         y^^ 
17  a'                ^ 

17  a\ 

lbac-{-37bc    , 
I0ab'—2c 

I0ab^2c. 

47am'-{-Sb-'C 

by 

ax""  —  3  a^ n 

ax^  —  Sa''m-{-b 

To  divide  a  fraction,     >        /7'  *//         ^the  numerator. 
To  multiply  a  fraction,  )  *^*  ^        (  the  denominator. 

Arith.  Art.  XVIII. 


Reducing  Fractiom  to  JLotver  Terms. 

XVII.  If  both  numerator  and  denominator  be  multiplied  by  tlie 
same  number,  the  value  of  the  fraction  will  not  be  altered. 

Arith.  Art.  XIX. 

For  multiplying  the  numerator  multiplies  the  fraction,  and 
multiplying  the  denominator  divides  it ;  hence  it  will  be  multi- 
plied and  the  product  divided  by  the  multiplier,  which  repro- 
duces the  multiplicand. 

In  other  words,  _?.  signifies  that  a  contains  b  a  certain  num- 
b 

ber  of  times,  if  a  is  as  large  or  larger  than  6  ;  or  a  part  of  one 
time,  if  b  is  larger  than  a.  Now  it  is  evident  that  2  a  will  con- 
tain 2  b  just  as  often,  since  both  numbers  are  twice  as  large  as 
before. 

So  dividing  both  numerator  and  denominator,  both  divides 
and  multiplies  by  the  same  number. 

3_2X3_6_7X3_21_3X6  __35 


2X5       10        7X5       35       5X&        56 
a  _2a       ba      ac      2acd 


b        2b       56       be       2bcd 
Gab  _Sb  X2a  _2a 
96c""36x3c~3c' 
8* 


W  Algeiyra.  XVII. 

Hence,  if  a  fraction  contain  the  same  factor  both  in  the  nu- 
merator and  denominator,  it  may  be  rejected  in  both,  that  is, 
both  may  be  divided  by  it.  This  is  called  reducing  fractions 
to  lower  terms. 

1.  Reduce  _ ^ —  to  its  lowest  terms.    Arts. 


Idbcm  5b  c 

2.  Reduce to  its  lowest  terms,     dns. 


16«'a?'  4  ax 

3.  Reduce  lli!^  to  its  lowest  terms.     Am.  ^^ 


30  bm  6b 

4.  Reduce  -^^^  ^  to  its  lowest  terms. 

5.  Reduce  to  its  lowest  terms. 

13a'¥x' 

6.  Reduce  ^A^lJlZIIL^L.  to  its  lowest  terms. 

5a'bc-{-d5a'b 

7.  Reduce 27  m^  a: —  54^ ^^  ^^^  j^^^^^^  ^^^^ 

lOSax'^Slx-^90m'x'' 

8.  Divide  35  a^  b  m^  x^     by     7  a^n  m^  x.  , 

Write  the  divisor  under  the  dividend  in  the  form  of  a  frac- 
tion, and  reduce  it  to  its  lowest  terms. 


9.  Divide     27  b'my'  by         21  b'  m'  y. 


Ans. 


a  n 
1  m. 


10. 

Divide     56  6  r'  y 

by 

lb'ny\ 

11. 

Divide     54  rn^nr^y 

by 

S6bmy\ 

12. 

Divide     IS  c^dmx^ 

by 

GScm'rx\ 

13. 

Divide  115  rsy' 

by 

Idrsy. 

14. 

Divide  128  a*  c*r  a:' 

by 

^Sa'mr'x', 

XVIII.  Addition  and  Subtraction  of  Fractions.  91 


15.  Divide 

17  acx 

by         \2ac^x\ 

16.  Divide 

2SaUy 

by         HttY- 

17.  Divide 

26  a' m^y 

by         54a^my\ 

18.  Divide 

Iba'bf 

by        ma^ey'x. 

19.  Divide 

a-f-6 

by         2c-.df. 

2a  Divide 

2a^c~7 

a'6c+  15«^c^ 

by         l^a'cd. 

21.  Divide 

18  a' m'- 

'■d4a'ni'  +  42a'm* 

by        30a'«i'(/— 12fl^cw\ 

22.  Di\ide{a-\'b)(lS 

23.  Divide  3c=(a— 2( 

acH-6c)   by   (m' —  c)  (a  +  J). 
0^     by     26c^(a  — 2c)^ 

24.  Divide  36  b'  c'{2a  +  df  (7  6  —  dy 

by     i2b'{2a-{-dY{7b-^dy{a^d), 

Mdition  and  Subtraction  of  Fractions, 

XVIII.     Add  together  ^  and  4  and  ±, 
b  d  f 

This  addition  may  be  expressed  by  writing  the  fractions  one 
after  the  other  w^ith  the  sign  of  addition  between  them ;  thus 

N.  B.  When  fractions  are  connected  by  the  signs  -f-  and 
— ,  the  sign  should  stand  directly  in  a  line  with  the  line  of  the 
fraction. 

It  is  frequently  necessary  to  add  the  numerators  together,  in 
which  case,  the  fractions,  if  they  are  not  of  the  same  denomi- 
nation, must  first  be  reduced  to  a  common  denominator,  as  in 
Arithmetic,  Art.  XIX. 

1.  Add  together -land  A.  Am.    i±^  =  i.. 

^  7  7  ^         ...  7  7 

2.  Add  together  ±  and  ±,  Ans,  i±5. 

b  b  b 


92  Algebra.  -  XVIII. 

3.  Add  toffether  _  and  —         Arts,   — ^ —  =  — . 

c  d        c  d  ^cd  c  d 

4.  Add  together  =-—  and  —iz.,  Ans.  — ^—~t- — 

^  Scd         2cd  Scd 

6.  Add  together   |   and   f . 

These  must  be  reduced  to  a  common  denominator.  It  has 
been  shown  above  that  if  both  numerator  and  denominator  be 
multiplied  by  the  same  number,  the  value  of  the  fraction  will 
not  be  altered.  If  both  the  numerator  and  denominator  of  the 
first  fraction  be  multiplied  by  7,  and  those  of  the  second  by  5, 
the  fractions  become  ||  and  if.  They  are  now  both  of  the 
same  denomination,  and  their  numerators  may  be  added.  The 
answer  is  f }. 

6.  Add  together  JL  and  .f. 

b  d 

Multiply  both  terms  ofvthe  first  by  d^  and  of  the  second  by 

6,  they  become  ^—.  and  ,_f .     The  denominators  are  now  alike 

b  d        0  d 
and  the  numerators  may  be  added. 

The  answer  is  ^L^_±lf. 
bd 

7.  Add  together  £.,    i.,    i_,  and    M-. 

0       d      f  h 

In  all  cases  the  denominators  vnll  be  alike  if  both  terms  of  each 
fraction  be  multiplied  by  the  denominators  of  all  the  others.  For 
then  they  will  all  consist  of  the  same  factors. 

Applying  this  rule  to  the  above  example,  the  fractions  be- 
come   "tllA,   i^,   ^llh.,  and  ^IL^-. 
bdfli   bdfh'   bdfli  bdfh 

The  answer  is  ^dfh  +  bcfh^bdeh+b  df  g 

bdfh 

8.  Add  together  i±  and  ^^    Ans.    l^l±t^. 

^  2bc         bd  \Obcd 


XVIII.  Addition  and  Subtraction  of  Fractions.  93 

It  was  shown  in  Arithmetic,  Art.  XXII,  that  a  common  de- 
nominator may  frequently  be  found  much  smaller  than  that  pro- 
duced by  the  above  rule.  This  is  much  more  easily  done  in 
algebra  than  in  arithmetic. 

9.  Add  together  JL.,    ,A,  and   L, 

h  (f      o  e  eg 

Here  the  denominators  will  be  alike,  if  each  be  multiplied 
by  all  the  factors  in  the  others  not  common  to  itself.  If  the 
first  be  multiplied  by  e  g,  the  second  by  c^  gy  and  the  third  by 
bee,  each  becomes  b  c^  e g.  Then  each  numerator  must  be 
multiplied  by  the  same  quantity  by  which  its  denominator  was 
multiplied,  that  the  value  of  the  fractions  may  not  be  altered- 

The  fractions  then  become  J?^,    -^lH,  and  ^4^^- 

be  eg    be  eg  cb^eg 

The  answer  is ''eg  +  ,fdg+bcef 
be^eg 

10.  Add  together  ?J^and?A/. 

^  be  2dg 

11.  Add  together  ^   ''  and^. 

2  r      36  h 

12.  Add  together  J^and   ^^ 


2m  n         Smp 

13.  Add  together  -4^-,   4^,  andiii.. 

*  dbrrv"     5bn  2m^n 

14.  Add  together  _ g._-,  and 


w'  r*  Smn^r 

15.  Add  together  lif-,    ^l^,and^^ 


2h'n''        h'n'x 
16.  Add  together  ?±£,  and  13  erf. 


17.  Add  together  ^±^,  and  2  a  c  —  5  6. 
4a  n 


94  Algebra.  XVIII. 

18.  Add  together  I3fl'/i^  — 4r  ^^  11  ac— 5w. 

2arn 

19.  Add  together    15 a'mc -2  a6 ^^  ga^c 

lamrb  36  m 

an    Ajj^       ^u       IS a^b  —  2c        J      lab  4- 8c 

20.  Add  together ,  and    — - — '       . 

^  4ab  2b  +  lQab 

21.  Subtract    —  from  — ,-• 

6c'  26c 

This  subtraction  may  be  expressed  thus, 
3a         e 


26c       6c' 

But  if  they  are  reduced  to  a  common  denominator,  the  nume- 
rators may  be  subtracted. 

Ans, 


Sac  —  2c 


2  6  c' 


22.  Subtract   ^   from  ^. 

23.  Subtract   -^SA-  from      "^^^ 


l¥m'x  21bm'af 

24.  Subtract  -^JL  from  ^''^ 


nx^y^  2m  y" 

^>y        fr^m         ^n 


25.  Subtract    -l^V  from 

26.  Subtract    11^^  from  li^JL. 

3m6  bnv'b^ 

27.  Subtract  11^  from  ILLI. 

28.  From         1 3  a  c  +  6  c  subtract  1?L£. 

26w 

29.  From        ?^^J^il5  subtract  1^5. 

2a^mx  14am 


XIX.  Division  hy  Fractions.  95 

21  ad     ,           2aJ(/  — 3m'c 
30.  From "^Yc^  subtract '^Y^ 

Solution. 

21ad_2ahd  —  ^cm^_    {21  ad)  2  h 
2he  4b'(f  (2  be')  2b 

_^(ih  d  —  3cwi^__    ^Aab  d 

2ab  d  —  3  c  wr  ^_  54  ab  d  —  2  a  b  d  +  S  c  m* 

4T'"?  4  b'  c^ 

__  52  ab  d'\'S  cm' 


4  b'  e 

which  is  the  answer. 

When  the  fraction  '^f^^d-^^cm'  ^^^    subtracted,  the 

4  b'  c' 
sign  —  was  changed  to  +.     See  Art.  VI,  example  6th. 

5  n  x'  —  \0  a  d  X 


31.  From 


Subtract 


\2ad 

l^nx''  —  5mx''-{-  17 
^  m  X 


32.  From       ^^  "  ^  ^      subtract 


^dx'  —  5  4dx 


\ 


XIX.     Division  of  whole  numbers  by  Fractions,  and  Fractions  by 

Fractions, 

1.  How  many  times  is  |  contained 'in  7? 

Ans,  I  is  contained  in  7,  35  times,  and  f  is  contained  ^  as 
many  times  ;  that  is,  \^  or  11|  times. 

2.  How  many  times  is  |  contained  in  a  ? 

Ans,  I  is  contained  in  a,  8  a  times,  and  f  is  contained  }  as 
many  times  ;  that  is,  Y  • 


96  Algebra,  XIX. 

3.  How  many  times  is  -^  contained  in  c  ? 

0 

Ans.     — -  is  contained  h  c  times  in  c,  and  -^  is  contained  _ 
o  ha 

he 
as  many  times ;  that  is,  — . 
a 

Arith.  Art.  XXIII. 

4.  Of  what  number  is  c  the  -^  part  ? 

h 

Ans,    If  c  is  the  ^  part  of  some  number,  —  will  be  -~- 

part  of  the  same  number,  and  £  is  _  part  of  ~ 

ah  a 

Arith.  Art.  XXIV. 

Hence,  to  divide  a  whole  number  by  ajraction^  multiply  it  by  the. 
denominator  of  the  fraction,  and  divide  the  product  by  the  numera- 
tor. 

How  many  times  is  |  contained  in  f . 

Solution,  Reducing  them  to  a  common  denominator,  |  is 
f  ^,  and  J  is  f  f .  f  f  is  contained  in  f  f  as  many  times  as  24  is 
contained  in  35  ;  that  is,  f  f  or  1^^,  Ami  l|i. 

6.  How  many  times  is  ~  contained  in  -£.  ? 

6  d 

Solution,    Reducing  them  to  a  conunon  denominator,  ^ 

0 

is  ?_ ,  and  4-  is  =-?•     ^,  is  contained  in  —  as  many  times  as 
bd  d      hd     bd  hd 

ad  19  contained  in  6  c ;  that  is,  —.  Ans.     —. 

ad  ad 

7.  Of  what  number  is  ~  the  ~  part  ? 

a  o 


XIX.  Division  of  Fractions.  97 

Solution.  If  -L  is  the  -^  part  of  some  number,  JL  part  of  — 
do  ad 

is  — .  part  of  that  nimiber ;  _  part  of  -I  is  — ,  and   —  is  -^ 
6  a  d        ad  ad       0 

part  of  ^.  ^n*.      *_i. 

Hence,  ^o  divide  a  fraction  by  a  fraction,  multiply  the  numerator 
of  the  dividend  by  the  denominator  of  the  divisor,  and  the  denomi- 
nator of  the  dividend  by  the  numerator  of  the  divisor. 

Or  more  generally,  when  the  divisor  is  a  fraction,  multiply  the 
dividend  (whether  whole  number  or  fraction)  by  the  divisor  inverted. 

Arith.  Arts.  XXIII.  and  XXIV. 


8.  Divide 

Sab 

by 

2 

9.  Divide 

IS  a 

by 

h 

c 

10.  Divide 

Ham 

by 

2e 
b' 

11.  Divide 

act 

by 

She 

2a 

12.  Divide 

Sax 

by 

2  a^m 
Sxy 

13.  Divide 

2ac--6 

c  by 

Sa 
5  c* 

14.  Divide 

17aa:'  — 

'2hx  + 

,       lSabx-2x 
^           7a'c 

15.  Divide 

11  ax'-- 

-Sx 

by            ^^         . 
7  acx — Sac 

16.  Divide 

he 
d 

by 

3  ac 

m 

17.  Divide 

2cd 
Say 

by 

2xy 
5ad^ 

18.  Divide 

17  a' m 

bx'  f 

9 

by 

Sa'n' 
7x'y'f 

99  Algebra.  XX. 

45a'n5  5 


19.  Divide 


35  6' m/    " 

13«  +  26c 

\2ax 

2a'-Scd 

2am  +  5ax 

5  mx  —  2d 

21  b'm'y 


20.  Divide       liiL±l^        by        13«ft-2«x' +7^ 

2  am  —  5  a  a? 
2a  +  3c6/  * 

13ay 


21.  Divide       -lii"""^'^"^        by 


22.  Divide       -^^-^^    by 


3my-^^md  bd  —  mx 

Division  of  Compound  Quantities, 

XX.  Sometimes  division  may  actually  be  performed  when 
both  divisor  and  dividend  are  compound  quantities.  Since  di- 
vision is  the  reverse  of  multiplication,  the  proper  method  to  dis- 
cover how  to  perform  it,  is  to  observe  how  a  product  is  formed 
by  multiplication. 

Multiply  2  a'  6  — •  3  a'  6'  c  +  a  6'  c' 
by  ^aH^  +  2ahc. 

8  a'  6'—  12  a*  J*  c  +  4  a' h'c"  +4a' 6' c  — 6  a'  V" c^  +  2a' b*c\ 

Observe  that  each  term  of  the  multiplier  is  multiplied  sepa- 
rately into  each  term  of  the  multiplicand.  The  product  there- 
fore must  consist  of  a  number  of  terms  equal  to  the  product  of 
the  number  of  terms  in  the  multiplicand  by  the  number  of  terms 
in  the  multiplier.  If  the  product  be  divided  by  the  multipli- 
cand, the  multiplier  must  be  reproduced,  and  if  by  the  multi- 
plier, the  multiplicand  must  be  reproduced. 

The  three  terms  8  a' b' —  12  a*  b*  c -{-4  a^  b'  (?  of  the  product 
were  produced  by  multiplying  the  three  terms  of  the  multipli- 
cand by  the  first  term  of  the  multiplier,  4  c^  6'.  Therefore,  if 
these  three  terms  be  divided  by  4  c^  6^,  the  quotient  will  be  the 
multiplicand. 

Again,  the  three  terms 

of  the  product  were  formed  by  multiplying  each  term  of  the 
multiplicand  by  2  a  6  c.  Therefore,  if  these  three  terms  be  di- 
vided by  2  a6  c,  the  quotient  will  be  the  multiplicand. 


XX.  Division  of  Compound  Quantities.  99 

Hence  we  see  that  the  whole  division  might  be  performed 
by  any  one  term  of  the  divisor,  if  all  the  terms  of  the  dividend 
which  depend  on  that  term  and  the  quotient  could  be  ascer- 
tained. This  cannot  often  be  done  by  inspection  ;  for  in  many 
products,  though  at  first  there  are  as  many  terms  as  there  are 
in  the  multiplicand  and  multiplier  together,  some  of  the  terms 
are  united  together  by  addition  or  subtraction,  and  some  disap- 
pear entirely.  Even  if  all  the  terms  did  remain  entire,  they 
could  not  be  easily  distinguished. 

However,  one  term  may  always  be  distinguished,  and  from  it 
one  term  of  the  quotient  may  be  obtained. 

Divide  4  a*  —  9  a*  6'  -j-  6  a  6'  •—  6*    , 

by  2a^-—Sab-\-b\ 

First,  it  is  evident  that  the  highest  power  of  either  letter  in 
the  dividend,  must  have  been  produced  by  multiplying  the  high- 
est power  of  that  letter  in  the  divisor  by  the  highest  power  of 
the  same  letter  in  the  quotient  ;  for  in  order  to  produce  the  di- 
vidend, each  term  of  the  divisor  must  be  multiplied  by  every 
term  of  the  quotient.  Therefore,  if  4  d^  be  divided  by  2  a^  it 
must  give  a  term  of  the  quotient.  Or,  if —  b*  be  divided  by  6^  it 
must  give  a  term  of  the  quotient.  Let  the  quantities  be  ar- 
ranged according  to  the  powers  of  the  letter  a. 

Dividend.  Divisor. 

4a*—9a'b'  +  6ab'-^ b*  /2  a' —-Sab -{- b' 
4a*'-6a'b  +  2a,'b'  \2a^-{-3ab -^¥  quotient. 


ea'b'-'9a'b'^Sab' 


^2a'b'  +  Sab'  —  b* 
—  2a'b'  +  Sab'-^b' 


I  divide  4  a^  by  2  a^  which  gives  2  a^  for  the  first  term  of  the 
quotient.  Now  in  forming  the  dividend,  every  term  of  the  di- 
visor was  multiplied  by  this  term  of  the  quotient,  therefore  I 


100  Algebra,  XX. 

multiply  the  divisor  by  this  term,  by  which  tneans  1  find  all  the 
terms  of  the  dividend,  vi^hich  depend  on  this  term.     They  are 

Here  is  a  term  6  a'  6  which  is  not  in  the  dividend,  this  must 
have  disappeared  in  the  product.  The  term  2  a^  V  is  not  found 
alone,  but  it  is  like  9  c^  V  and  must  have  disappeared  by  unit- 
ing with  some  other  term  to  form  that.  I  subtract  these  three 
terms  from  the  dividend,  and  there  remains 

which  does  not  depend  at  all  on  the  term  2  a*  of  the  quotient, 
but  which  was  formed  by  multiplying  each  remaining  term  of 
the  quotient  by  all  the  terms  of  the  divisor.  This  then  is  anew 
dividend,  and  to  find  the  next  term  of  the  quotient  we  must 
proceed  exactly  as  before  ;  that  is,  divide  the  term  of  the  divi- 
dend containing  the  highest  power  of  a,  which  is  6  a^  5,  by  2  o* 
of  the  divisor,  because  this  must  have  been  formed  by  multiply- 
ing 2  €?  by  the  highest  remaining  power  of  a  in  the  quotient. 
This  gives  for  the  quotient  -|-  3  a  6.  I  multiply  each  term  of 
the  divisor  by  this,  and  subtract  the  product  as  before,  and  for 
the  same  reason.     The  remainder  is 

—  2a'^6  +  3a6^--5S 

which  depends  only  on  the  remaining  part  of  the  quotient. 
The  highest  power  of  «,  viz.  2  c^  h^,  must  have  been  produced 
by  multiplying  some  term  of  the  quotient  by  2  a^  of  the  divisor ; 
therefore  I  divide  by  this  again,  and  obtain  —  h^  for  the  quo- 
tient. I  multiply  by  this  and  subtract  as  before,  and  there  is 
no  remainder,  which  shows  that  the  division  is  completed. 

By  the  above  process  I  have  been  enabled  to  discover  all 
the  terms  of  the  dividend  produced  by  multiplying  the  first 
term  of  the  divisor  by  each  term  of  the  quotient.  If  both  be 
arranged  according  to  the  powers  of  the  letter  6,  and  the  same 
course  pursued,  the  same  quotient  will  be  obtained,  but  in  a 
reversed  order. 

In  the  division  the  term  —  2  a'  &^  has  the  sign  — .  Here  we 
must  observe  that  the  divisor  and  quotient  multiplied  together 
must  reproduce  the  dividend. 

If  -f-  a  J  be  divided  by  +  a,  the  quotient  must  be  -|-  6,  be- 
cause -}-  a  X  +  6  gives  ^  ah. 


XX.        „  Division  of  Compound  Quantities,  101 

-    If — ab  be  divided  by  -\-  a,  the  quotient  must  be  —  6,  be- 
cause +  a  X  —  b gives  —  ab. 

If  -f-  «  6  be  divided  by  —  a,  the  quotient  must  be  —  6,  be- 
cause —  a  X  —  b  gives  -\-  ab. 

If — a  5  be  divided  by  — a,  the  quotient  must  be  -}-  b,  be- 
cause —  a  X  -\-b  gives  —  ab. 

The  rule  for  signs  therefore  is  the  same  as  in  multiplication. 

WTien  the  signs  are  alike,  that  is,  both  -\-  or  both  — ,  tJie  sign  of 
the  product  must  be  +  ;  but  when  the  signs  are  unlike,  that  is,  one 
-f-  and  the  other  — ,  the  sign  of  the  quotient  must  be  — . 

By  the  reasoning  above  we  derive  the  following  rule  for  di- 
vision of  compound  numbers. 

Arrange  the  dividend  and  divisor  according  to  the  powers  of 
some  letter.  Divide  the  first  term  of  the  dividend  by  tJie  first  term 
of  the  divisor,  and  write  the  result  in  the  quotient.  Multiply  all  the 
terms  oftlie  divisor  by  the  term  of  the  quotient  thus  found,  and  sub- 
tract the  product  from  the  dividend.  The  remainder  will  be  a  new 
dividend,  and  in  order  to  find  the  next  term  of  the  quotient,  proceed 
exactly  as  before  ;  and  so  on  until  there  is  no  remainder. 

Sometimes,  however,  there  will  be  a  remainder,  such  that 
the  first  term  of  the  divisor,  will  not  divide  either  term  of  it ;  in 
which  case  the  division  can  be  continued  no  farther,  and  the 
remainder  must  be  written  over  the  divisor  in  the  form  of  a  frac- 
tion, and  annexed  to  the  quotient  as  in  arithmetic. 

Divide  2  a'—  11  a*  b -{- M  a'  b^ -\- 13  a'  b'  by  2  a— 6. 

r2a-^b 


2 a' ^  lla*b  -{-n  a' b'  +  ISa' b' 
2a'-~aH     ~ 

—  10a*6-f-lla'6^+13a^6^ 

—  10a*&-l-5a^6« 

a'^5a'b-^Sa'b'+Sab' 
L                 2a  —  b 

6a^V  —  3a'b' 

16a»6' 

8ab' 
Sab'-- 

-4b' 

4J^ 


lOe  Algtbra,  XXI. 

In  this  example,  the  division  may  be  continued  until  the  re- 
mainder is  4  h^y  which  cannot  be  divided  by  «,  therefore  it  must 
be  written  over  the  divisor  2  a  —  6  as  a  fraction  and  added  to 
the  quotient. 

Examples, 

1.  Divide  a?*  -}-  2  a  a:  4*  «'  by  a  +  a?.* 

2.  Divide  a*  —  6'  by  a-j-ft. 

3.  Divide  h'  -\-2h^x  +  x'  by  h^  +  x, 

4.  Divide  x^  —  y^  by  a?®  +  ^  y  +  y'* 

5.  Divide  o?^  —  y*  by  ^  +  y. 

6.  Divide  ISa'^-f  2a6  — 8i'  by  ^a  —  2L 

7.  Divide  a:'  —  2a:y*  +  y'  by  ^ — V- 

8.  Divide  a'— 9a'-}- 27  by  9  — 6a-f  a'. 

9.  Divide  4rt*  — 3  —  9  a'  -^  6  a 

by       3  cc  —  1  +  2  a^ 

10.  Divide  a*  —  x*  by       a*  —  a^  x  -{-  a  .r'  —  a:*. 

11.  Divide  6a?'  — 96  by       3a:  — 6. 

12.  Divide  4a''  —  ab  by       2a  —  6.    . 

13.  Divide  6  a*  H-9  a'-— 15  a     by     3  a'  — 3a. 

XXI.     Equations. 

The  above  rules  are  sufficient  to  solve  all  equations  of  the 
first  degree. 

1.  Find  the  value  of  a?  in  the  equation 

ah^  X  —  2  c         2  at  ,  h*  x 

— =  a  0  0?  — . 

5  a  3  a — h  3 

First,  clear  it  of  fractions  by  multiplying  by  the  denomina- 
tors. 

*  Let  the  learner  prove  his  results  by  multiplication. 


}■■-■. 


XXI.  Equations.  -    109 

;    Expressing  the  multiplication,  we  have 

(a5»a:  — 2c)  (3«--6)  (3)  — (2  a  c)  (5  a)  (3) 
=  (a6^)(5a)(3a  — 6)(3)-.(6'a:)(5a)(3a  — J). 

Performing  the  multiplication  it  becomes 

9  a' 6' a?—  18  ac  —  ^a¥x  +  ^hc  —  30  a*  c 
z=zAba?hx---lbaH^x-^lbaH''x+bah'x. 

Transposing  all  the  terms  which  contain  x  into  the  first  mem- 
ber, and  those  which  do  not  contain  it  into  the  second  member, 
it  becomes 

9a2  i^a?  —  3  fl6*  a:  —  45  a'  6  x-\- 15  a^h^  a?  + 15  a*  6*a?—  5  «  6't 
=;  18«c — 66c-f  30a*c. 

Uniting  the  terms  which  are  alike 

39a*5'a:— -8a6'a?  — 45a'&a?=  18ac— 66c  +  30fl*r. 
Separating  the  first  member  into  factors 
(39a9  52  — 8a6'  — 45a'5)a?  =  18ac-— 66c-|-30a'c, 

,.  .     .  18ac — 6ic  +  30a''c 

which  mves  x  = — — -!- — -. 

^  39«*6*  — 8a6»  — 45a'^> 

2.  Find  the  value  of  a?  in  the  following  equation ; 

13a  — ^  =  2ca?-|-rf. 
2c  ^ 

3.  What  is  the  value  of  a:  in  the  following  equation  ? 


h  —  3a?  3a6 — 126c 

4.  What  is  the  value  of  a?  in  the  following  equation? 

^^       ~136c  =  ^'^~^ 


5a?  — 2a  26  —  1 

6.  What  is  the  vahie  of  a:  in  the  following  equation  ? 


36c — 2aa:  1  —  56 


10*  mgehra.  XXII. 

XXII.     MisceUaneom  Examples  producing  Simple  Equations* 

1.  A  merchant  sent  a  venture  to  sea  and  lost  one  fourth  of  it 
by  shipwreck ;  he  then  added  $2250  to  what  remained,  and 
sent  again.  This  time  he  lost  one  third  of  what  he  sent.  He 
then  added  $1000  to  what  remained,  and  sent  a  third  time,  and 
gained  a  sum  equal  to  twice  the  third  venture  ;  his  whole  re- 
turn was  equal  to  three  times  his  first  venture.  What  was  the 
value  of  the  first  venture  ^ 

2.  A  man  let  out  a  certain  sum  of  money  at  6  per  cent,  sim- 
ple interest,  which  interest  in  10  years  wanted  but  £12  to  be 
equal  to  the  principal.     What  was  the  principal  ? 

3.  A  man  let  out  £98  in  two  different  parcels,  one  at  5,  and 
the  other  at  6  per  cent,  simple  interest ;  and  the  interest  of  the 
whole,  in  15  years,  amounted  to  £81.  What  were  the  two  par- 
cels ^ 

4.  A  shepherd  driving  a  flock  of  sheep  in  time  of  war,  met  a 
company  of  soldiers,  who  plundered  him  of  one  half  the  sheep 
he  had  and  half  a  sheep  over  ;  the  same  treatment  he  received 
from  a  second,  a  third,  and  a  fourth  company,  each  succeeding 
company  plundering  him  of  one  half  the  sheep  he  had  left  and 
one  half  a  sheep  over.  At  last  he  had  only  7  sheep  left.  How 
many  had  he  at  first  r* 

6.  A  man  being  asked  how  many  teeth  he  had  remaining, 
answered,  three  times  as  many  as  he  had  lost ;  and  being  asked 
how  many  he  had  lost,  answered,  as  many  as,  being  multiplied 
into  I  part  of  the  number  he  had  left,  would  give  the  number 
he  had  at  first.  How  many  had  he  remaining,  and  how  many 
had  he  lost  ? 

After  this  question  is  put  into  equation  every  term  may  be 
divided  by  x. 

6.  There  is  a  rectangular  field  whose  length  is  to  its  breadth 
as  3  to  2,  and  the  number  of  square  rods  in  the  field  is  equal  to 
i5  times  the  number  of  rods  round  it.  Required  the  length  and 
breadth  of  the  field. 

7.  What  two  numbers  are  those,  whose  difference,  sum,  and 
product,  are  to  each  other,  as  the  numbers  2,  3,  and  5  respec- 
tively ? 


XXII.  Miscellaneous  Examples,  105 

8.  Generalize  the  above  by  putting  a,  b,  and  c  instead  of  2, 
3,  and  5  respectively. 

Let  X  =z  the  greater 
and  y  =  the 

Then 

1. 


2. 

3.  by  the  first 

4.  by  the  2nd 

5.  by  3d  and  4th 

6.  dividing  by  x 

ax-^-c      6  +  a 

7.  clearing  of  fractions  be  -fac=.abx  —  a*  x-^b  c  —  ac 

8.  by  transposition         abx  —  a* xz=:2ac 

9.  from  the  8th  (J  —  a)  ^  =  2  c. 

2c 


cc  — y 

a 

X  — 

yz=-^xy 

y-'- 

X  —  ax  __ 
6-f  a 

6  +  0 

ex 

^       ax-\-c 

ex 

_b-^a 
b-^a 

.  X 

ax-\-c 

c      _b 

—  a 

b^a 

b  —  a         2c  2c 


10.  x=z 

11.  putting  10th  into  3d  y  =  

6  -}-  a       6  —  a      b  -\-  a 

Solve  the  7th  Ex.  by  these  formulas ;  also  try  other  num- 
bers. 

9.  When  a  company  at  a  tavern  came  to  pay  their  reckon- 
ing, they  found  that  if  there  had  been  three  persons  more,  they 
would  have  had  a  shilling  apiece  less  to  pay ;  and  if  there  had 
been  two  less,  they  would  have  had  to  pay  a  shilling  apiece 
more.  How  many  persons  were  there,  and  how  much  had 
each  to  pay  ? 


106  Algebra.  XXII. 

10.  A  sum  of  money  is  to  be  divided  equally  among  a  cer- 
tain number  of  persons.  Now  if  there  were  three  claimants 
less,  each  would  receive  150  dollars  more;  and  if  there  Were 
6  more,  each  would  receive  120  dollars  less.  How  many  per- 
sons are  there,  and  how  much  is  each  to  receive  ? 

11.  What  fraction  is  that,  to  the  numerator  of  which  if  1  be 
added,  its  value  will  be  \,  but  if  1  be  added  to  its  denominator 
its  value  will  be  \. 

12.  What  fraction  is  that,  to  the  numerator  of  which  if  a  be 

added,  its  value  will  be  ^'  but  if  a  be  added  to  its  denomina- 
n 

tor  its  value  will  be-Z.? 

Arts.     Numerator    ^^(^+!^, 
mq  —  np 

Denominator   ""(y+g). 
m  q  —  n  p 

Solve  the  11th  example  by  these  formulas. 

13.  What  fraction  is  that,  from  the  numerator  of  which  if  a 

be  subtracted,  its  value  will  be   — ,  but  if  a  be  subtracted  from 

n 

its  denominator,  its  value  will  be  ^  ? 

q 

N.  B.  The  answers  to  the  12th  and  13th  differ  only  in  the 
signs  of  the  denominators.  The  learner  will  find  by  endeavour- 
ing to  solve  particular  examples  from  these  formulas,  that  he 
will  not  always  succeed.  If  in  making  examples  for  the  12th, 
he  selects  his  numbers,  so  that  n  ph  greater  than  m  q,  the  for- 
mula will  fail  ;  but  if  he  takes  the  same  numbers,  and  applies 
them  according  to  the  conditions  of  the  13th,  they  will  answer 
those  conditions.  When  m  9  is  greater  than  n  p  the  numbers 
will  not  suit  the  conditions  of  the  13th,  but  they  will  answer  to 
those  of  the  12th.  The  numbers  in  example  11th  will  not  form 
an  example  according  to  the  13th.  The  following  numbers 
will  form  an  example  for  the  I3th  but  not  for  the  12th. 


XXIII.  Equations,  107 

14.  What  fraction  is  that,  from  the  numerator  of  which  if 
3  be  subtracted,  its  value  will  be  f ,  but  if  3  be  subtracted  from 
its  denominator  its  value  will  be  y^  ^ 

The  reason  why  numbers  chosen  indiscriminately  will  not 
satisfy  the  conditions  of  the  above  formulas  will  be  explained 
hereafter. 

Equations  with  several  Unknoum  Quantities. 

XXIII.     Questions  involving  more  than  two  unknown   Quan- 
tities, 

Sometimes  it  is  necessary  to  employ,  in  the  solution  of  a 
question,  more  than  two  unknown  quantities.  In  this  case,  the 
question  must  furnish  conditions  enough  to  form  as  many  dis- 
tinct equations  as  there  are  unknown  quantities. 

1.  A  market  woman  sold  to  one  man,  7  apples,  10  pears, 
and  12  peaches,  for  63  cents  ;  and  to  another,  13  apples,  6 
pears,  and  2  peaches,  for  31  cents  ;  and  to  a  third,  1 1  apples, 
14  pears,  and  8  peaches  for  63  cents.  She  sold  them  each 
time  at  the  same  rate.     What  was  the  price  of  each  ? 

Let  X  =z  the  price  of  an  apple, 
y  =  "  a  pear, 

z  =i  "  a  peach. 

Then  we  shall  have 

1.  7  a?  +  10  2/  +  12  z  =  63 

2.  13  a?  +    6  y  -f    2  z  =  31 

3.  11  cc  +  14y  -{-    8z  =  63. 

The  second  being  multiplied  by  6,  the  z  will  have  the  same 
coefficient  as  in  the  first. 

4.  78a?-|-36y+ 12z  =  186 
I.  7a?+10y-f-12z=    63 

5.  71  a? +  26  3/       *      =:  123. 

If  the  second  be  multiplied  by  4,  the  z  will  have  the  same 
coefficient  as  the  3d. 


108  Algebra,  XXIII. 

6.  62a?  +  24y  +  8^=  124  •     * 
3.                          lla:+14y-|-8z=    63 

7.  41a:+10j^       *    =    61 

We  have  now  the  two  equations       71a:  +  26y=:123 
and  41a?+  10y=   61 

which  contain  only  two  unknown  quantities.  These  may  now 
be  reduced  in  the  same  manner  as  others  with  two  unknown 
quantities. 

Muhiplying  the  5th  by  5,  and  the  7th  by  13,  the  coefficient 
of  y  will  be  the  same  in  both. 

8.  355  a; -I-  130y  =  615 

9.  533  a?  +  130^  =  793 


10.  178  a?         *      =  178 

We  have  now  found  an  equation  containing  only  one  un- 
known quantity. 

178  a:  =178 

X=z        1. 

Putting  the  value  of  x  into  the  7th,  it  becomes 

41  -{-10y  =  61 

10y  =  20 

y-   2, 

Putting  the  values  of  x  and  y  into  the  2d,  it  becomes 

13 -f- 12  +  2^=31 
2:2=    6 
«=    3. 

Ans.  The  apples  1,  the  pears  2.  and  the  peaches  3  cents 
each. 

In  the  same  manner,  questions,  involving  four  unknown  quan- 
tities, may  be  solved.  First  combine  them  two  by  two  till  one 
of  the  unknown  quantities  is  eliminated  from  the  whole,  and 
there  will  be  three  equations  with  three  unknown  quantities. 
Then  combine  these  three  two  by  two,  until  one  of  the  un- 


e 


XXIII.  Equations.  100 

known  quantities  is  eliminated,  and  then  there  will  be  two 
equations  with  two  unknown  quantities,  and  so  on. 

Either  of  the  methods  of  elimination  may  be  used  as  is  most 
convenient. 

It  is  not  necessary  that  all  the  unknown  quantities  should 
enter  into  every  equation. 

2.  A  market  woman  sold  at  one  time  7  eggs,  12  apples,  and 
a  pie  for  26  cents  ;  at  another  time  12  eggs,  18  pears,  and  3 
pies,  for  69  cents  ;  at  a  third  time  20  pears,  10  apples,  and  17 
eggs  for  69  cents  ;  and  at  a  fourth  time,  7  pies,  18  apples,  and 
10  pears  for  66  cents.  Each  article  was  sold,  at  every  sale,  at 
the  same  price  as  at  first.  What  was  the  price  of  each  ar- 
ticle ? 

Let  u  =  the  price  of  an  egg, 
X  z=  "  an  apple, 

y  =  "  a  pie, 

z  =■  "  a  pear. 

1.  7M+12a?+       3/ =26 

2.  12w-f-18  2+    3y  =  69 

3.  17  M  +  20  z  + 10  a?  =  69 

4.  10z-fl8a?+   7y=:66 

5.  In  the  1st,  ^=26  —  1  u  —  \2  x. 

Putting  this  value  of  y  into  the  2nd  and  4th,  they  become 

6.  12w+ 18^  +  78— 21w  —  36a?  =  69 

7.  10;r  +  18j:+ 182  — 49m  — 84a?  =  66.  -    ^ 

Uniting  and  transposing  terms 

8.  18z—    9m  —  36a;= —     9 

9.  lOz  — 49m  — 66^  =  — 116 
3.             20z-j- 17M4-10a?=         69 

If  the  9th  be  multiplied  by  2,  the  coefficient  of  z  will  be  the 
same  as  in  the  3d; 

10.  20  ;^  —  98  M  —  132  X  =  —  232. 

10 


O 


110  Algebra,  XXIU. 

Subtracting  10th  from  3d 
3.  20z+  17w+    10a:  =  69 

10.  20«  — 98m— 1320?  =  — 232 


11.*  *       115M+ 142cc=:301 

If  the  8th  be  multiplied  by  5,  and  the  9th  by  9,  the  coeffi- 
cients of  z  will  be  alike. 

12.  9O2:—    45m— 180^  =  — 45 

13.  90  z  —  44 1  M  —  594  a:  =  —  1044. 
Subtracting  13th  from  12th 

14.  396  m  4- 414  a?  =  999. 

Deducing  the  value  of  a?  from  11th,  and  also  from  14th. 
301  — 115  m 


15. 


16.  xz=i 


142 

999  — 396  m 


414 

Making  these  values  of  x  equal,  we  have  an  equation  con- 
taining only  one  unknown  quantity. 

999  — 396m  _  301 — 115m 

414  142 

This  equation  solved  in  the  usual  way  gives 

M  =  2 

Putting  this  value  of  m  into  the  15th  or  16th,  we  shall  find 

1 

a:  =- 
2' 

Putting  these  values  of  x  and  u  into  the  1st,  2nd,  or  4tli,  and 
we  shall  find 

Putting  the  values  of  a?  and  u  into  the  3d,  and  we  shall  find 

z—  11. 

Ans,  Eggs,  2  cents  each,  cipples,  ^  cent,  pears,  1|  cent,  and 
pies,  6  cents. 

*  If  the  learner  is  at  a  loss  how  to  subtract  — 233  from  69  let  him  transpose 


XXm.  Equations,  Ul. 

In  this  example,  three  different  methods  of  elimination  were 
employed.  This  was  not  necessary  ;  either  method  might  have 
been  used  for  the  whole.  It  is  sometimes  convenient  to  use 
one,  and  sometimes  the  other. 

3.  There  are  three  persons.  A,  B,  and  C,  whose  ages  are  as 
follows ;  if  B's  age  be  subtracted  from  A's,  the  difference  will 
be  C's  age  ;  if  five  times  B's  age  and  twice  C's  age  be  added 
together,  and  from  their  sum  A's  age  be  subtracted,  the  re- 
mainder will  be  147  ;  the  sum  of  all  their  ages  is  96.  What 
are  their  ages  ? 

4.  Three  men,  A,  B,  C,  driving  their  sheep  to  market, 
says  A  to  B  and  C,  if  each  of  you  will  give  me  5  of  your  sheep, 
I  shall  have  just  half  as  many  as  both  of  you  will  hr^e  left. 
Says  B  to  A  and  C,  if  each  of  you  will  give  me  5  of  yours,  I 
shall  have  just  as  many  as  both  of  you  will  have  left.  Says  C 
to  A  and  B,  if  each  of  you  will  give  me  5  of  yours,  I  shall  have 
just  twice  as  many  as  both  of  you  will  have  left.  How  many 
had  each .'' 

5.  It  is  required  to  divide  the  number  72  into  four  such 
parts,  that  if  the  first  part  be  increased  by  5,  the  second  part 
diminished  by  5,  the  third  part  multiplied  by  5,  and  the  fourth 
part  divided  by  5,  the  sum,  difference,  product,  and  quotient, 
shall  all  be  equal. 

6.  A  grocer  had  four  kinds  of  wine,  marked  A,  B,  C,  and  D. 
He  mixed  together  7  gallons  of  A,  5  gallons  of  B,  and  8  gal- 
lons of  C,  and  sold  the  mixture  at  $1.21  per  gallon.  He,  also 
mixed  together  3  gallons  of  A,  10  of  C,  and  5  of  D,  and  sold 
the  mixture  at  $1.50  per  gallon.  At  another  time  he  mixed  8 
gallons  of  A,  10  of  B,  10  of  C,  and  7  of  D,  and  sold  the  whole 
for  $48.  At  another  time  he  mixed  together  18  gallons  of  A, 
and  15  of  D,  and  sold  the  mixture  for  $48.  What  was  the 
value  of  each  kind  of  wine  ^ 

7.  Find  the  values  of  w,  x,  y,  and  z,  in  the  following  equa- 
tTons. 

X  —  2y  -{-  Sz  ^=z  5u 
Sx — 15 —  u  =  4y  —  23 
2w  -|-  z  —  y  =  27 
y  _j-  12  — 3a?-f  11m=91. 


112  Algebra.  XXIV. 

S.  Three  persons,  A,  B,  and  C,  talking  of  their  money, 
says  A  to  B  and  C,  give  me  half  of  your  money  and  I  shall 
have  a  sum  d ;  says  B  to  A  and  C,  give  me  one  third  of  your 
money  and  I  shall  have  d;  says  C  to  A  and  B,  give  me  one 
fourth  of  your  money,  and  I  shall  have  d.  How  much  had 
each  ? 


XXIV.    JVegative  Quantities, 

It  sometimes  happens  in  the  course  of  a  calculation,  through 
some  misconception  of  the  conditions  of  the  question,  that  a 
quantity  is  added  which  ought  to  have  been  subtracted,  or  a 
quantity  subtracted  which  ought  to  have  been  added.  In 
this  case,  algebra  will  detect  the  error,  and  show  how  to 
correct  it. 

The  length  of  a  certain  field  is  a,  and  its  breadth  I ;  how 
much  must  be  added  to  its  length,  that  its  content  may  be  c  .'^ 

Let  X  z=i  the  quantity  to  be  added  to  the  length. 
Then  a  -\-  x  z=z  the  length  after  adding  x. 
ah  -\-hx  =z  c 

hx=z  c  —  ab 

X  =  -^ — a. 
6 

Suppose  the  length  to  be  8  rods,  and  the  breadth  6  ;  how 
much  must  be  added  to  the  length,  that  the  field  may  contain 
60  square  rods  f 

Here  a  =  8, 6  =  5,  and  c  =  60 

60  Q  . 

a?  =  —  —  8  =  4. 
5 

Ans,  4  rods,  and  the  whole  length  will  be  12  rods. 

Suppose  the  length  8  rods,  and  the  breadth  5  ;  how  much 
must  be  added  to  the  length,  that  the  field  may  contain  30 
square  rods  f 

5 


XXIV.  JVegative  Quantities.  113 

The  answer  is  —  2  rods.  What  shall  we  understand  by 
this  negative  sign  ? 

Let  us  return  to  the  original  equation.  .  . 

8x54-507  =  30 
or  40  +  5  a;  n:  30. 

Here  appears  an  absurdity  in  supposing  something  to  be 
added  to  40  to  make  30.  The  result  shows  that  we  must  add 
—  2  rods,  that  is,  subtract  2  rods,  which  is  in  fact  the  C€ise  ; 
for 

40  —  5  X  2=:  30. 

Let  the  question  be  proposed  as  follows.  There  is  a  field  8 
rods  long  and  5  wide  ;  how  much  must  be  subtracted  from  the 
length,  that  the  field  may  contain  30  square  rods  ? 

40  —  5  JT  =:  30 
oc=    2. 

The  value  ofx  is  now  positive,  which  shows  that  the  ques- 
tion is  correctly  expressed. 

There  is  a  field  8  rods  long  and  5  rods  wide,  how  much 
must  be  subtracted  from  the  length,  that  the  field  may  contain 
50  square  rods  ?  * 

40  —  5  a:  =  50 

x  =  —  2. 

/  Here  again  the  value  of  a?  is  negative,  which  shows  some  in- 
consistency in  the  question. 

The  inconsistency  consists  in  supposing  that  something  must 
be  subtracted  from  40  to  make  50.  In  order  to  correct  it,  sup- 
pose something  added.  That  is,  put  into  the  equation  -\-  5  x 
instead  of  —  5  x. 

Hitherto  we  have  treated  of  nogative  quantities  only  in  con- 
nexion with  positive.  They  arise  from  the  necessity  of  express- 
ing subtraction  by  a  sign,  because  it  cannot  actually  be  per- 
formed on  dissimilar  quantities.  They  are  only  positive  quan- 
tities subtracted,  and  in  their  nature  they  differ  in  nothing  from 
positive  quantities.  In  that  connexion  we  discovered  rules  for 
operating  upon  the  quantities  affected  with  the  sign  — . 

It  may  sometimes  happen  as  we -have  just  seen,  that  by  some 
wrong  supposition  in  the  conditions  of  the  question,  the  quan- 
tities to  be  subtracted  may  become  greater  than  those  from 
10* 


114  Algebra.  XXIV. 

which  they  are  to  be  subtracted,  in  which  case  the  whole  ex- 
pression taken  together,  or  which  is  the  same  thing,  the  result 
after  subtraction,  will  be  negative.  This  is  what  is  properly 
called  a  negative  quantity. 

A  negative  quantity  cannot  in  reality  be  a  quantity  less  than 
nothing,  but  it  implies  some  contradiction.  It  answers  to  a 
figure  of  speech  frequently  used.  If  it  is  asked,  how  much  a 
man  is  worth  who  owes  five  thousand  dollars  more  than  he  can 
pay,  we  sometimes  say  he  is  worth  five  thousand  dollars  less 
than  nothing,  instead  of  changing  the  form  of  expression  and 
saying,  he  owes  five  thousand  dollars  more  than  he  can  pay. 

If  any  thing  is  added  to  a  number,  properly  speaking  it  must 
increase  the  number  ;  if  we  add  nothing,  it  is  not  altered.  It 
is  impossible  to  add  less  than  nothing  ;  but  by  a  figure  of 
speech  we  may  use  the  expression,  add  a  quantity  less  than  no- 
thing, to  signify  subtraction. 

As  these  negative  quantities  may  frequently  occur,  it  is  ne- 
cessary to  find  rules  for  using  them. 

In  the  first  place,  let  us  observe,  that  all  negative  quantities 
are  derived  from  endeavouring  to  subtract  a  larger  quantity 
from  a  smaller  one.  The  largest  number  that  can  actually  be 
subtracted  from  any  number,  is  the  number  itself.  Thus  the 
largest  number  that  can  be  subtracted  from  5  is  5  ;  the  largest 
number  that  can  be  subtracted  from  a  is  a  itself.  If  it  be  re- 
quired to  subtract  8  from  5,  it  becomes  5  —  5  — 3  =  —  3  ;  the 
5  only  can  be  subtracted,  the  3  remains  with  the  sign  — ,  which 
shows  that  it  could  not  be  subtracted.  If  5  be  subtracted  from 
8,  the  remainder  is  3,  the  same  as  in  the  other  case  except  the 
sign. 

In  the  same  manner,  if  it  be  required  to  subtract  b  from  a,  h 
being  the  larger  the  remainder  will  have  the  sign  — ,  that  is, 
a  — b  will  be  a  negative  quantity. 

Suppose  b  —  azzzm-,  then  a  —  b=.  —  m.  That  is,  whether 
a  be  subtracted  from  b  or  b  from  «,  the  numerical  value  of  the 
remainder  is  the  same,  differing  only  with  respect  to  the  sign. 

It  is  required  to  add  the  quantity  a  —  6  to  r . 

The  answer  is  evidently  c  -f-  «  —  b. 

Now  if  a  is  greater  than  b,  the  quantity  c  -\-  a  —  6,  is  greater 
than  c,  by  the  difference  between  a  and  6  ;  but  if  b  is  greater 
than  a,  the  quantity  is  smaller  than  c,  by  the  difference  between 
a  and  b.     That  is,  if 


XXIV.  Negative  Quantities.  118 


b  —  a  =  w, 

then 

a  — b  =  —  m 

and 

c  -{-  a  —  b  =  c  —  m. 

Hence,  adding  a  negative  quantity,  is  equivalent  to  subtract 
ing  an  equal  positive  quantity. 

In  the  above  example  of  the  field,  in  which  the  length  was  8 
rods  and  breadth  5,  it  was  asked,  how  much  must  be  added  to 
the  length,  that  it  might  contain  30  square  rods.  The  answer 
was  —  2  ;  which  was  equivalent  to  saying,  you  must  subtract 
2  rods. 

It  is  required  to  subtract  a  —  b  from  c. 

The  answer  is  evidently  c  —  a  -}-  b. 

Now  if  a  is  greater  than  &,  the  quantity  c  —  a  -[-bis  less  than 
c  by  the  difference  between  a  and  b,  but  if  b  is  greater  than  a, 
the  quantity  is  larger  than  c,  by  the  same  quantity. 

Let     a  —  b  =:  —  m     which  gives     —  a  -{-b:=zm 

then    c  —  a  -{- b  ^=  c  -\- m. 

Hence,  subtracting  a  negative  quantity,  is  equivalent  to 
adding  an  equal  positive  quantity. 

In  the  example  of  the  field,  in  which  the  length  was  8  rods 
and  the  breadth  5,  it  was  asked,  how  much  must  be  subtracted 
from  the  length,  that  the  field  might  contain  50  square  rods. 

The  answer  was  —  2  rods,  which  was  equivalent  to  saying 
that  2  rods  must  be  added  to  the  length. 

A  is  worth  a  number  a  of  dollars,  B  is  not  worth  fo  much  as 
A  by  a  number  b  of  dollars,  and  C  is  worth  c  times  as  much  as 
B.     How  much  is  C  worth  ? 

B's  property     =  a  — -b. 
C's  property     =z  a  c  —  be 

Now  if  a  is  greater  than  &,  the  quantity  ac  —  b  c  will  be  po- 
sitive ;  but  if  b  is  greater  than  «,  then  a  —  &  is  negative,  and 
also  ac  —  6  c  is  negative. 

Let  b  —  a=.m. 

then  be  —  acz=icm. 

and  ac  —  bc-=.  —  cm. 

or  c{a  —  6)  =  —  cm. 


11^  Algdra.  XXIV. 

That  is,  if  B  is  in  debt,  C  is  c  times  as  much  in  debt.  Hence 
if  a  negative  quantity  be  multiplied  by  a  positive,  the  product 
is  negative. 

A  gentleman  owned  a  number  a  of  farms,  and  each  farm  was 
worth  a  number  c  of  dollars,  which  was  his  whole  property. 
He  hired  money  and  fitted  out  a  number  h  of  vessels,  and  each 
vessel  was  worth  as  much  as  one  of  his  farms.  All  the  vessels 
were  lost  at  sea.     How  much  was  he  then  worth. 

He  was  worth  a  —  h  times  c  dollars.  That  is,  a  c  —  he  dol- 
lars. 

Now  if  the  number  of  farms  exceeded  the  number  of  vessels, 
he  still  had  some  property,  but  if  the  number  of  vessels  exceed- 
ed the  number  of  farms,  (that  is,  if  h  is  larger  than  a,)  the 
quantity  a  c  —  6  c  is  negative,  and  he  owed  more  than  he  could 
pay. 

Hence  if  a  positive  quantity  be  multiplied  by  a  negative  the 
product  will  be  negative. 

Multiply     a  —  h  by  c  —  d, 

a  —  b 
c-d        ^ 


Product  ac  —  he  —  ad  ^h  d. 

This  product  may  be  put  in  this  form. 

(a  —  h)c  -\-{h  —  a)  d. 

Let  it  be  remembered  that  a  —  h  has  the  same  numerical 
value  as  6  —  a,  they  differ  only  in  the  sign. 

Suppose  a  —  h  z=z  — »wj 

by  changing  all  the  signs  h  —  a  =  -{-m. 

Hence  (a  —  h)e-\-  {h  —  a)d=z  —  em-^dm=:m{d  —  c). 

Now  if  d  is  greater  than  c,  (which  is  the  case  when  c  —  rf  is 
negative,)  the  quantity  m  {d  —  c)  is  positive. 

Hence  if  a  negative  quantity  be  multiplied  by  a  negative, 
the  product  will  be  positive. 

Another  demonstration.  Suppose  both  a  —  h  and  c  —  d  to 
be  negative,  as  before  ;  then  h  —  a  and  d  —  c  will  both  be  po- 
sitive, and  their  product  will  be  positive. 


XXIV..  JVegative  Quantities.  \\7 

b  —  a 
d  —  c 


hd  —  he  —  ad-\-ac. 

This  product  is  precisely  the  same  as  that  produced  by  mul- 
tiplying a  —  6  by  c  —  d.  Therefore  if  two  negative  quantities 
be  multiplied  together,  the  product  will  be  the  same  as  that  of 
two  positive  quantities  of  the  same  numerical  value,  and  will 
have  the  positive  sign. 

,It  is  required  to  find  the  second  power  of  a — 6,  and  also  of 
h  —  a. 

The  second  power  of  each  is  a*  -j-  5*  —  2  ah. 

Now  if  a  — h  is  positive,  then  h  —  a  is  negative  ;  or  if  a  — h 
is  negative,  then  h  —  a  is  positive. 

Suppose  a  —  h  =:m  , 

then  h  —  a  =  —  m 

we  have  («  — -  by  =  {b  —  «)*  =  m*. 

That  is,  the  second  power  of  any  quantity,  whether  positive 
or  negative,  is  necessarily  positive. 

The  rules  for  division  will  necessarily  follow  from  those  of 
multiplication. 

Hence  the  rules  which  apply  to  terms  affected  with  the  sign 
—  in  compound  quantities,  extend  to  isolated  negative  quan- 
tities. 

We  might  also  derive  the  same  rules  in  the  following  man- 
ner. It  has  been  shown  that  a  negative  quantity  is  derived 
from  some  contradiction  in  the  conditions  of  question,  by  which 
that  quantity  entered  into  the  equation  with  the  wrong  sign. 
Now,  in  order  to  make  it  right,  the  sign  of  that  quantity  must 
be  changed  in  all  places  where  it  is  used.  That  is,  if  it  was 
before  added,  it  must  now  be  subtracted  ;  and  if  it  was  sub- 
tracted before,  it  must  now  be  added,  and  that  whether  multi- 
plied by  another  quantity  or  not. 

Suppose  we  have  the  equation 

ax  —  2  J?'  —  2ahx  =  0  —  rf. 

Now  suppose  that  we  have  x  =  —  m. 


118  Algebra,  XXIV. 

This  shows  that  x  was  used  in  all  cases  with  the  wrong  sign, 
therefore  to  insert  —  m  in  place  of  x  we  must  change  the  sign 
in  each  term  where  x  is  found. 

Take  the  quantity  first  without  x,  thus, 
a  — 2  —  2<ib. 

First  insert  — w  in  the  second  term  and  it  becomes 

a -{-2  m  —  2ab, 
Now  insert  —  m  into  all  the  terms,  and  it  becomes 
—  am  —  2m*4-2a6m=:c  —  d. 

If  —  m  be  inserted  by  the  rules  found  above,  the  same  re- 
sult will  be  produced. 

When  a  negative  value  has  been  found  for  the  unknown 
quantity,  we  have  observed  it  shows  that  there  was  some  in- 
consistency in  the  question.  If  then  the  unknown  quantity  be 
put  again  into  the  same  equation,  with  the  contrary  sign,  as  we 
introduced  —  m  above,  that  is,  if  the  unknown  quantity  be  ta- 
ken with  the  negative  sign,  and  introduced  by  the  above  rules 
into  all  the  terms  where  it  was  found  before,  a  new  equation 
will  be  produced,  diftering  from  the  former  only  in  some  of  the 
signs.  Then  if  the  conditions  of  the  question  be  altered  so  as 
to  correspond  with  the  new  equation,  it  will  be  consistent,  and 
a  positive  value  will  be  obtained  for  the  unknown  quantity. 
The  new  value  of  the  unknown  quantity  however  will  be  the 
same  as  the  former,  with  the  exception  of  the  sign.  There- 
fore, when  once  we  are  accustomed  to  interpret  this  kind  of 
results,  it  will  be  unnecessary  to  go  through  the  calculation  a 
second  time. 

The  following  examples  are  intended  to  exercise  the  learner 
in  interpreting  these  results. 

1.  A  father  is  55  years  old,  and  his  son  is  16.  In  how  many 
years  will  the  son  be  one  fourth  as  old  as  the  father  ? 

Let  X  =  the  number  of  years. 

16 -f-    X  = — -C- 
4 

64  +  4  a?  =  55  +  a; 

3a:=:55  — 64  =  --9 

a:  =  —  3. 


XXIV.  J^egative  Quantities,  119 

Here  x  has  a  negative  value,  consequently  it  entered  into 
the  equation  with  the  wrong  sign.  Putting  now  —  x  instead 
of  a:  into  the  equation,  it  becomes 

-^  55  —  X,    ' 

16  —  X  =z  

4 

This  shows  that  something  must  be  subtracted  from  the  pre- 
sent age ;  that  is,  the  son  was  a  fourth  part  as  old  as  the  father 
some  years  before. 

This  equation  gives 

X=zS. 

Therefore  he  was  one  fourth  part  as  old  3  years  before, 
when  the  father  was  52,  and  the  son  13. 

2.  A  man  when  he  was  married  was  45  years  old,  and  his 
wife  20.     How  many  years  before,  was  he  twice  as  old  as  she  .'* 

,^^  45  —  X 

20  —  X  = 

2 

x  =  —  5. 

There  is  a  wrong  supposition  in  this  question.  Putting  —  x 
into  the  equation  it  becomes 

20  4-x  =z   — X — 

^  2 

a?  =  5. 

This  shows  that  she  was  not  half  as  old  as  he  when  they  were 
married,  but  that  it  was  to  happen  5  years  afterward,  when  the 
man  was  50,  and  the  wife  25. 

3.  A  labourer  wrought  for  a  man  15  days,  and  had  his  wife 
and  son  with  him  the  first  9  days,  and  received  $14.25.  He 
afterwards  wrought  12  days,  having  his  wife  and  son  with  him 
5  days,  and  received  $13.50.  How  much  did  he  receive  per 
day  himself,  and  how  much  for  his  wife  and  son  ? 

4.  A  labourer  wrought  for  a  man  1 1  days,  and  had  his  wife 
with  him  4  days,  and  received  $17.82.  He  afterwards  wrought 
23  days,  having  his  wife  with  him  13  days,  and  received  $38.78. 
How  much  did  he  receive  per  day  for  himself,  and  how  much 
for  his  wife  ? 


120  Algebra.        '■  XXIV. 

5.  A  labourer  wrought  for  a  gentleman  7  days,  having  his 
wife  with  him  4  days,  and  his  son  3  days,  and  received  $7.89. 
At  another  time  he  wrought  10  days,  having  his  wife  with  him 
7  days,  and  his  son  5  days,  and  received  $11. G5.  At  a  third 
time  he  wrought  8  days,  having  his  wife  with  him  5  days,  and 
his  son  8  days,  and  received  $7.54.  How  much  did  he  re- 
ceive per  day  himself,  and  how  much  for  his  wife  and  son  se- 
verally ? 

6.  What  number  is  that,  whose  fourth  part  exceeds  its  third 
part  by  16.? 

?=f4-16 
4       3^ 

0?  =  — 192. 

The  question  as  it  was  proposed  involves  some  contradiction. 
Putting  in  —  a?  it  becomes 

—  £  =  —.?-[-  16. 
4  3^ 

Changing  all  the  signs 

4  "~  3"~ 

X  =   192. 

This  shows  that  the  question  should  have  been  as  follows  ; 
What  number  is  that,  whose  third  part  exceeds  its  fourth  part 
by  16  ? 

7.  What  number  is  that,  j\  of  which  exceeds  4  of  it  by  18  ? 

8.  What  fraction  is  that,  to  the  numerator  of  which  if  1  be 
added,  its  value  will  be  |,  but  if  1  be  added  to  its  denominator, 
its  value  will  be  4  .'* 

9.  What  fraction  is  that,  from  the  denominator  of  which,  if 
2  be  subtracted,  its  value  will  be  |a,  but  if  2  be  subtracted 
from  its  numerator,  its  value  will  be  f  ? 

10.  It  is  required  to  divide  the  number  20  into  two  such 
parts,  that  if  the  larger  be  multiplied  by  3,  and  the  smaller  by 
5,  the  sum  of  the  products  will  be  125. 


XXV.  \egatim  Exponents.  121 

11.  It  is  required  to  find  two  numbers,  whose  difference  is 
25,  and  such  that  if  the  larger  be  multiplied  by  7,  and  the 
smaller  by  5,  the  difference  of  their  products  shall  be  215? 

XXV.     Explanation  of  Negative  Exponents. 

It  was  observed  above,  that  when  the  dividend  and  the  divi- 
sor were  different  powers  of  the  same  letter,  division  is  per- 
formed by  subtracting  the  exponent  of  the  divisor  from  that  of 
the  dividend  :  thus 

Now  ^  =  1 .     By  the  above  principle  -  =  a'^'  =  a° ;  there- 
a  a 

foreft*'^  1. 

Also  ^  :==«='-='  =  ««  =  1  ;  ^  =  6->  =  6«  =  1  ; 


=  10'-' =10*^=1;  -!L±A  ^  (a -I- 6) '"' 


a"  h 

10 
10  '    a^h 

^{aJ^hy^l^ 

That  is,  any  quantity  having  zero  for  its  exponent,  is  equal 
to  1. 

Again  ^  =  1,  or   ^  =  a'-'  z=z  a^' 
fl'        a  a* 


Hence  it  appears  that  a"'  has  the  same  value  as  1,  and  a" 

a 
1 


as 
a* 

The  quantities  a\  a*,  a\  a\  «"',  «-*,  a-\  &c.  have  the  same 

value  as  a%  a',  a',  1,  -,  _,  _,  (fee 

a   a*    a^ 

*  Exponents  may  be  used  for  compound  quantities  as  well  as  for  simple ; 
and  multiplication  and  division  may  be  performed  o>i  those  which  are  similar, 
by  adding  and  subtracting  the  exponents. 
11 


122  Algebra.  XXV. 

On  this  principle  the  denominator  of  a  fraction,  or  any  fac- 
tor of  the  denominator  may  be  written  in  the  numerator  by  giv- 
ing its  exponent  the  sign — .  This  mode  of  notation  is-often 
very  convenient ;  I  shall  therefore  give  a  few  examples  of  its 
application. 

2  a        2ab~^ 


be 


=  2ab-"  c-\ 


1.  Multiply      ^     by        b'c. 

By  the  common  rule  l^xb^c^    2abU  _   2ab' 


u  V  b  c^  ( 

By  the  principle  explained  above, 
2ab-'c~'  X  b'c=2ab-'-^'c-^-^'=2ab'c-'  =  ^' 


2.  Multiply  3  a  c  -'  ^  -'     by     3  a*  c'  d\ 

3.  Multiply  da-'  C-'     by     2a c\ 

4.  Multiply  li^     by     3a^c^ 


5. 

Multiply  2  a 

(b+d) 

—3 

by 

3 

a(b^  d)\ 

6. 

Multiply   — 
4  c 

Sab 
(2  a— 6 

^[ 

by 

8  c' 

'(2a  — 6< 

7. 

Divide  ^f 

by 

c' 

"^"^  '' 

.^: 

By 

the  common 

I  method 

Sa 

~c' 

'  = 

3a 
c'* 

By 

the  above  method  Sac" 

'-J-C 

8   — 

:3ac-^~' 

=:Sac- 

-6   

3a 

c" 

Or  thus,  to  divide  3  a  c  "^  by  c',  is  the  same  as  to  multiply  it 

by  —  or  c  """,  which  gives  the  same  result, 
c* 


f,. 


XXVI.  Examination  of  General  Formulas,  123 

8.  Divide   ,//"-,    by     a'(2  6-.c)«. 

(2  6  —  cy 

9.  Multiply?:?         by        44- 

c  a      .  c  a 

l^=z3ac-'d~'  and -11.=  4ec-'(^"". 
cd  c"  d 


c'dJ" 


10.  Divide  ?-?-       by  ^^ 


ed  "  a'c'd 

2ac~'d-'  ^Sba-'c-'d-' 

2a'+''b-'  c-'-^'d-'  +  '        2a'b'''       2a' 


3  3  Sb 

In  this  example  the  exponents  to  be  subtracted  had  the 
sign  — ,  which  in  subtracting  was  changed  to  -{-. 

11.  Multiply    ^^P-^y    by  __I^L!!L_. 

^^  2bc  ^    3{bc  —  dY 

12.  Multiply   -^^^'     ,       by      ^lA^^ZpTj 

^^    3  c  (6— 2  c)'         ^  12a' b' 

13.  Divide  l^Jbc-^)'  by     ^^«^(^^-^)\ 

16  6=^  c^  ^  24  6^  c^ 

14.  Divide  ^^0^^  +  ^^)^    by    5a-(176  +  3^r 

(a_-6y  ^         4  (a  — 6)^ 

XXVI.     Examination  of  General  Formulas. 

When  a  question  has  been  resolved  generally,  that  is,  by^  re- 
presenting the  known  quantities  by  letters,  we  sometimes  pro- 
pose to  determine  what  values  the  unknown  quantities,  will 
take,  for  particular  suppositions  made  upon  the  known  quan- 
tities. 


124  Algebra.  XXVI. 

The  two  following  questions  offer  nearly  all  the  circuni- 
stances  that  can  ever  occur  in  equations  of  the  first  degree. 

A C B 

Two  couriers  set  out  at  the  same  time  from  the  points  A  and 
B,  distant  from  each  other  a  number  m  of  miles,  and  travel 
towards  each  other  until  they  meet.  The  courier  who  sets 
out  from  the  point  A,  travels  at  the  rate  of  a  miles  per  hour ; 
the  other  travels  at  the  rate  of  6  miles  per  hour.  At  what  dis- 
tance from  the  points  A  and  B  will  they  meet  } 

Suppose  C  to  be  the  point,  and 

Let  X  =  the  distance  A  C 

and  y  =  the  distance  B  C. 

For  the  first  equation  we  have 

X  -\-  y  =1  K  V^  z=z  m 

Since  the  first  courier  travels  x  miles,  at  the  rate  of  a  miles 

per  hour,  he  will  be  ~  hours  upon  the  road.     The  second  cou- 
a 

rier  will  be  \.  hours  upon  the  road.      But  they  travel  equal 

0 

times;  therefore, 

^  _  X 
a         h 

Putting  this  value  of  a?  into  the  first  equation,  it  becomes 
-^-hy^m 

ay-^by^z  bm 
bm 


y- 


a  +  b 


__  ay  __    a        bm    __       abm      __    am 


b        a-\-b        b{a-A^b)         «-}.& 

Since  neither  of  the  quantities  in  these  values  of  x  and  y  has 
the  sign  — ,  it  is  impossible  for  cither  value  to  become  nega- 


XXVI.  Examination  of  Formulas.  125 

tive.  Therefore  whatever  numbers  may  be  put  in  place  of  a, 
h,  and  m,  they  will  give  an  answer  according  to  the  conditions 
of  the  question.  In  fact,  since  they  travel  towards  each  other, 
whatever  be  the  distance  of  the  places,  and  at  whatever  rate 
they  travel,  they  must  necessarily  meet. 

Suppose  now  that  the  two  couriers  setting  out  from  the  points 
A  and  B  situated  as  before,  both  travel  in  the  same  direction 
towards  D,  at  the  same  rates  as  before.  At  what  distances 
from  the  points  A  and  B  will  the  place  of  their  meeting,  C, 
be? 

A  BCD 


Let  X  =  the  distance  from  A  to  C, 
and  y  =  "  B  to  C. 

X  — yrrAC— BC  =  AB  =  Wl. 

The  second  equation  expressing  only  the  equality  of  the 


time  will  not  be  altered. 


a      h 


Solving  the  two  equations  as  before, 

ay 


^^v^ 


m 


ay  —  hy  =  bm 

b  m 

y~ X 

a — 0 

a  a      bm      abm       am 

b  b    a — b         b  (a  —  b)         a  —  6* 

Here  the  values  of  x  and  y  will  not  he  positive  unless  a  is 
greater  than  b  ;  that  is,  unless  the  courier,  that  sets  out  from  A, 
travels  faster  than  the  other. 

Suppose  a  =  8  and  6=4. 
11* 


126  Mgehra.  XXVI. 

Then  x=-^_  =:  ^=2//* 

8  —  4  4 

^       8  —  4  4 

In  this  case  the  point  C,  where  they  come  together,  is  dis- 
tant from  A  twice  the  distance  A  B. 

Suppose  a  smaller  than  h,  for  example 

0  =  4  and  6  =:  8. 

Then  x=.A^^^^m 

4  —  8 

^       4  —  8 

Here  the  values  of  x  and  y  are  both  negative ;  hence  there 
IS  some  absurdity  in  the  enunciation  of  the  question  for  these 
numbers.  In  fact,  it  is  impossible  that  the  courier  setting  out 
from  A,  and  travelling  slower  than  the  other  should  overtake 
him. 

Let  us  put  X  and  y  negative  in  the  two  equations,  that  is, 
change  their  signs. 


They  become 

—  ^  "f"  y  =^  ^« 

1 

a             h 

or 

y  —  a:  =  m 

and 

^  -.  y 

a            6' 

The  second  equation  is  not  affected  by  changing  the  sign  ; 
and  it  ought  not  to  be  so,  since  it  expresses  only  the  equality  of 
the  times. 

The  first  equation  becomes  y  —  x  =zm^  instead  of  x  —  y  zzz 
m,  which  shows  that  the  point  where  they  are  together  is  nearer 
to  A  than  to  B,  by  the  distance  from  A  to  B.  It  must  therefore 
be  on  the  other  side  of  A,  as  at  E. 

E  A  B  C  D 


XXVI.  Examination  of  Formulas.  127 

The  enunciation  of  the  question  may  be  changed  in  two 
ways  so  as  to  answer  the  conditions  of  this  equation. 

First,  we  may  suppose,  that  the  couriers,  setting  out  from  A 
and  B,  instead  of  going  towards  D,  go  in  the  opposite  direction, 
the  one  from  A  at  4  miles  per  hour,  and  the  other  from  B  at  8 
miles  per  hour  ;  at  what  distance  from  the  points  A  and  B  is 
the  point  E,  where  they  come  together  ? 

Or  we  may  suppose  that  two  couriers  setting  out  from  the 
same  place  E,  one  travelling  at  the  rate  of  4  miles,  and  the  other 
8  per  hour,  have  arrived  at  the  same  time  at  the  points  A  and 
B,  which  are  m  miles  asunder.  What  distance  are  the  points 
A  and  B  from  E  ? 

Suppose  a  =zb. 


Then  x  = 


a  in  am  am 


—  5         a  ■— «  0 

bm  h  m.         am 


a  —  6        a  —  a         0 

How  is  this  result  to  be  interpreted  ? 

Observe  that  in  this  case  a  and  b  being  equal,  the  two  cou- 
riers travel  equally  fast,  it  is  therefore  impossible  that  one 
should  ever  overtake  the  other,  however  far  they  may  travel  in 
either  direction,  and  no  change  in  the  conditions  can  make  it 
possible.     Zero  being  divisor,  then,  is  a  sign  of  impossibility. 

We  may  observe  that  when  there  is  any  difference,  however 
small,  between  a  and  5,  the  values  of  x  and  y  will  be  real,  and 
the  couriers  will  come  together  in  one  direction  or  the  other ; 
and  the  smaller  the  difference,  the  greater  will  be  the  distance 
travelled  before  they  come  together ;  that  is,  the  greater  will 
be  the  values  of  o^  and  y. 

Suppose         a  =:  5  and  b  =z  4,     a  —  6  =  1, 

then  x=^^:=5m     i,  =  t^  =  4m. 

1  ^1 

Again,  Suppose  «  =  5,  and  6  =  4-  5,  a  —  6  =  - 5, 

then  x=:tl!L=:lOm    y=i:i^=9m. 

•5  ^  -5 


128  Algebra,  XXVL 

Again,  Suppose  a  =  5,  and  Z>  =  4  •  98,  a  —  6  =  •  02, 

then         x  =  ^=  260m,  and  y  =  ^-L??  =  249  tw- 
•02  ^        -02 

Again,  Suppose  a  =  5  and  b  =  4*998,  a  —  b  =  -002, 

then  a?  =  Ajl  =  2500  w. 

•002 

and  y=t2^^=.2499m, 

^  2 

Here  observe,  that  as  the  difference  between  a  and  b  be- 
comes very  small,  the  values  of  a;  and  y  become  very  large,  and 
the  difference  between  them  is  always  m.  Hence,  since  the 
smaller  the  divisor  the  larger  the  quotient,  we  may  conclude, 
that  when  the  divisor  is  actually  zero,  the  quotient  must  be  in- 
finite.    From  this  consideration,  mathematicians  have  called 

the  expression-,  that  is,  a  quantity  divided  by  zero,  a  symbol 

of  infinity.  They  therefore  say,  that,  both  couriers  travelling 
equally  fast,  the  distance,  travelled  before  they  come  together, 
is  infinite.  But  as  infinity  is  an  impossible  quantity,  I  prefer 
the  term  impossible^  as  being  a  term  more  easily  comprehended. 

I  shall  therefore  call  -  a  symbol  of  impossibility. 

If  a  quantity  be  divided  by  an  infinite  or  impossible  quan- 
tity, the  quotient  will  be  zero.     If  b  be  divided  by  -,  it  be- 
comes —     Multiplv  both  numerator  and  denominator  by  0,  it 
a 

0 

becomes  — li.-  =  0.     In  fact,  since  the  larger  the  divisor,  the 
a 

smaller  the  quotient,  the  dividend  remaining  the  same,  it  fol- 
lows that  if  the  divisor  surpasses  any  assignable  quantity,  the 
quotient  must  be  smaller  than  any  assignable  quantity,  or  no- 
thing. 

One  case  more  deserves  our  notice.  It  is  when  a  =z  b  and 
»i  =  0 ;  in  which  case  we  have 


XXVL                    Examination  of  Formulas.  l£9 

am    a  X  0   0 


X  = 


a — h 


If  we  return  to  the  equations  themselves,  they  become 
X  —  y  =i  0 

a         a  .  , 

From  the  first  we  have  X 

a:  =  y 
Substituting  this  value  in  the  second 

a         a 

This  last  equation  has  both  its  members  alike,  and  is  some- 
times called  an  identical  equation.  The  values  of  the  unknown 
quantities  cannot  be  determined  from  it.  In  fact,  since  m  is 
zero,  both  couriers  set  out  from  the  same  point.  And  since 
they  both  travel  at  the  same  rate,  they  are  always  together. 
Therefore  there  is  no  point  where  they  can  be  said  to  come  to- 
gether. The  expression  -  is  here  an  expression  of  an  indeter- 
minate quantity. 

There  are  some  cases  where  an  expression  of  this  kind  is  not 
a  sign  of  an  indeterminate  quantity,  but  in  these  cases  it  arises 
from  a  factor  being  common  to  the  numerator  and  denomina- 
tor, which  by  some  suppositions  becomes  zero,  and  renders  the 

fraction  of  the  form  of  _  ;  but  being  freed  from  that  factor,  it 
has  a  determinate  value. 
The  following  expression  is  ajn  example  of  it. 

«K--6') 
6  (a  — 6)' 

When  a  =  6,  this  expression  becomes  -.     But  both  nuroera- 


130  Algebra.  XXVI. 

rator  and  denominator  contain  the  factor  a  —  h,  which  be- 
comes zero  when  a  and  h  are  equal. 

Dividing  by  a  — ft,  the  expression  becomes 

which  is  equal  to  2  a  when  a  =  b. 

It  is  necessary  then,  when  we  find  an  expression  of  the  fonn 
^-,  before  pronouncing  it  an  indeterminate  quantity,  to  see  if 

there  is  not  a  factor,  common  to  the  numerator  and  denomina- 
tor, which,  becoming  zero,  renders  the  expression  of  this  form. 

The  example  of  the  couriers  furnishes  some  other  curious 
cases,  for  which  we  must  refer  the  learner  to  Lacroix's  or  Bour- 
don's Algebra. 

Let  the  learner  examine  the  following  examples  in  a  similar 
manner. 

In  Art.  IX.  examples  15  and  16,  the  following  formulas,  re- 
lating to  interest,  were  obtained.  How  are  r  and  t  to  be  in- 
terpreted, when  p  is  greater  than  « ;  and  how  when  a  and  p 
are  equal? 

tp  rp 

In  Art.  XXII.  examples  12th  and  13th,  the  following  formu- 
las were  obtained.  In  what  cases  will  the  results  become  ne- 
gative, and  how  are  the  negative  results  to  be  interpreted  ^ 

12th.  Numerator    «H?^±^) 
mq  —  np 

Denominator  J!!liJP-I±3K 
mq  —  np) 

13th.  Numerator    "^('"+^ 
np  —  mq 

Denominator  ^^<E±i).' 
np  —  mq) 


XXVII.  Equations  of  the  Second  Degree,  ISl 

It  is  required  to  divide  a  given  number  a  into  two  such  parts, 
that  if  r  times  one  part  be  added  to  s  times  the  other  part,  the 
sum  will  be  a  given  number  b. 

'  Ans.  The  part  to  be  multiplied  by  r  is         ^^, 

r  —  s 

and  the  part  to  be  multiplied  by  s  is  — HI — 

r —  5 

In  what  cases  will  one  or  both  of  these  results  be  negative  ? 
Can  both  be  negative  at  the  same  time  ?  How  are  the  nega- 
tive results  to  be  interpreted  ?  In  what  cases  will  either  of 
them  become  zero  ?  Can  both  become  zero  at  the  same  time  ? 
What  is  to  be  understood  when  one  or  both  become  zero  ?  In 
what  cases,  will  one  or  both  become  infinite  or  impossible  ? 

Can  either  of  them  ever  be  of  the  form  _  ? 


XXVII.     Equations  of  the  Second  Degree. 

1.  A  boy  being  asked  how  many  chickens  he  had,  answered, 
that  if  the  number  were  multiplied  by  four  times  itself,  the  pro- 
duct would  be  256.     How  many  had  he  t 

Let  X  =  the  number, 

then  Ax  zzz  four  times  the  number. 

4a:  X  a:  =  4x* 
By  the  conditions         4  a?*  =  256 

X^=1^4: 

That  is  a?  a:  =  64. 

This  equation  is  essentially  different  from  any  which  we  have 
hitherto  seen. 

It  is  called  an,  equation  of  the  second  degree,  because  it  con- 
tains x^,  or  the  second  power  of  the  unknown  quantity.  In  or- 
der to  find  the  value  of  x,  it  is  necessary  to  find  what  number, 
multiplied  by  itself,  will  produce  64.  We  know  immediately 
by  the  table  of  Pythagoras  that  8  X  8  =  64.     Therefore 

X  :=.  %.  Ans,  8  chickens, 

JVote,  The  results  of  these  equations  may  be  proved  like 
those  of  the  first  degree. 


138  Algebra.  XXVIL 

2.  A  boy  being  asked  his  age,  answered,  that  if  it  were  mul- 
tiphed  by  itself,  and  from  the  product  37  were  subtracted,  and 
the  remainder  multiplied  by  his  age,  the  product  would  be  12 
times  his  age.     What  was  his  age  ? 

.T  X  a?  =  CF*     (x»  —  37)  X  =  a?'—  37  r. 

By  the  conditions 

x^— 37a?=:12x. 

Dividing  by  a:, 

x*  — 37  =  12 


a?  =    7.  Am.  7  years. 

3.  There  are  two  numbers  in  the  proportion  of  5  to  4,  and 
the  difference  of  whose  second  powers  is  9.  What  are  the  num- 
bers ? 

Let  X  =  the  larger  number, 

then  —  =  the  smaller. 
5 

4 13?  •    1 6  J? 

The  second  power  of  —  is ^ 

^  5         25 

16  cr* 

By  the  conditions  x^  — =9. 

^  25 

4.  There  are  two  numbers  whose  sum  is  to  the  less  in  the 
proportion  of  1 5  to  4,  and  whose  sum  multiplied  by  the  less 
produces  135.     What  are  the  numbers  ? 

Let  X  =  the  less,  and  y  =  the  greater,  y 

Then    x  -\~y=^ 

and  JJ  (a?  -|-  y)  =  135. 

rru              J    •                135  — a?" 
The  second  gives  y  = 

Putting  this  value  of  y  into  the  first,  it  becomes 

,   135  — x*         15a?  jf 
z  + =  ,_-,&c. 

X  4  '  1  . 


XXVIII.  Extraction  of  the  Second  Root.  lo.i 

Hence  it  appears,  that  when  an  example  involves  the  second 
power  of  the  unknown  quantity,  the  value  of  the  second  power 
must  first  be  found  in  the  same  manner  as  the  unknown  quan- 
tity is  found  in  pimple  equations  ;  and  from  the  value  of  the 
second  power,  the  value  of  the  first  power  is  derived. 

It  is  easy  to  find  the  second  power  of  any  quantity,  when 
the  first  power  is  known,  because  it  is  done  by  multiplication  ; 
but  it  is  not  so  easy  to  find  the  first  power  from  the  second.  It 
cannot  be  done  by  division,  because  there  is  no  divisor  given. 
When  the  number  is  the  second  power  of  a  small  number,  tht^ 
first  power  is  easily  found  by  trial,  as  in  the  above  examples. 
When  the  number  is  large,  it  is  still  found  by  trial ;  but  a  rule 
may  be  very  easily  found,  by  which  the  number  of  trials  will 
"  be  reduced  to  very  few.  The  first  power  is  called  the  root  of 
the  second  power,  and  when  it  is  required  to  find  the  first  pow- 
er from  the  second,  the  process  is  called  extracting  the  root. 

It  has  been  shown,  Art.  XXIV.  that  the  second  power  of 
every  quantity,  whether  positive  or  negative,  is  necessarily  po- 
sitive ;  thus  3  X  3  =  +  9,  and  also  — 3X— 3=:-f-9.  So 
«  X  «  —  a^  and  also  —  a  X  —  fl  =  «*.  Hence  every  second 
power,  properly  speaking,  has  two  roots,  the  one  positive  and 
the  other  negative.  The  conditions  of  the  question  will  gene- 
rally show  which  is  the  true  answer. 


XXVIII.     Extraction  of  the  Second  Root. 

In  order  to  find  a  rule  for  extracting  the  root,  or  finding  the 
first  power  from  the  second,  it  will  be  necessary,  first,  to  ob- 
serve how  the  second  power  is  formed  from  the  first. 

Let  a  =z20  and  h  =:  1  -,  then  a  -|-  6  =  27. 
The  second  power  of  a  -|-  6  is  ' 

(a  +  6)  (a  -h  6)  =:  a'  -h  2  a  6  -f  U'. 
a'  =  20  X  20  —  400 
ab  =z  20  X  7=  140 
«^;  —  20  X  7  —  140 
6^  —  7  X  7  =  49 
.  a-"  -{-2ah  -^b'  —  729. 
12 


r 


134  Algebra.  XXVIII. 

The  product  is  formed  in  precisely  the  same  manner  in  the 
usual  mode  of  multiplication,  as  may  be  seen,  if  the  products 
are  written  down  as  they  are  formed,  without  carrying. 

27 
27 

49 
140 

140 
400 


729 

Here  we  observe,  7  times  7  is  49,  7  times  20  is  140, 20  times 
7  is  140,  and  lastly  20  times  20  is  400.  These  added  together 
make  729,  which  is  the  second  power  of  27. 

We  ob^rve, 

1st.  When  the  root  or  first  power  consists  of  two  figures,  the 
second  power  consists  of  the  second  power  of  the  tens,  plus  the 
product  of  twice  the  tens  by  the  units,  plus  the  second  power 
of  the  units. 

2d.  The  second  power  of  9,  the  largest  number  consisting 
of  one  figure,  is  81  ;  and  the  second  power  of  10,  the  smallest 
number  consisting  of  two  places,  is  100  ;  and  the  second  pow- 
er of  100,  the  smallest  number  consisting  of  three  places,  is 
1 0000.  Hence,  when  the  root  consists  of  one  figure,  the  se- 
cond power  cannot  exceed  two  figures  ;  and  when  the  root 
consists  of  two  figures,  the  second  power  consists  of  not  less 
than  three  figures,  nor  more  than  four  figures. 

From  these  remarks  it  appears,  that  we  must  first  endeavour 
to  find  the  second  power  of  the  tens,  and  that  it  will  be  found 
among  the  hundreds  and  thousands. 

Let  it  be  required  to  find  the  root  of  729.  This  number  con- 
tains hundreds,  therefore  the  root  will  contain  tens.  The  se- 
cond power  of  the  tens  is  contained  in  the  700.  20  X  20  is 
400,  and  30  X  30  is  900.  400  is  the  greatest  second  power 
of  tens  contained  in  700.  The  root  of  400  is  20.  Subtract 
400  from  729,  and  the  remainder  is  329.  This  must  contain  - 
2  a  6  -|-  ^^  that  is,  the  product  of  twice  the  tens  by  the  units, 
plus  the  second  power  of  the  units.     If  it  contained  exactly  the 


XXVIII.  Extraction  of  the  Second  Root.  135 

product  2  a  6  of  twice  the  tens  by  the  units,  the  units  of  the 
root  would  be  found  by  dividing  329  by  twice  20,  or  40 ;  for 
2  ah  divided  by  2  a  gives  b.  As  it  is,  if  we  divide  by  twice  20 
or  40,  we  shall  obtain  a  quotient  either  exact,  or  too  large  by  1 
or  2.  40  is  contained  in  329,  8  times.  Write  8  in  the  root 
and  raise  the  whole  to  the  second  power.  28  X  28  =  784, 
which  is  larger  than  729.  Next  try  7  in  the  place  of  8.  27  X 
27  —  729.     Therefore  7  is  right,  and  27  is  the  root  required. 

The  operation  may  stand  as  follows. 

729  (20  -I-  7  =  27  root. 
400 


329  (40  divisor. 

27  X  27 

=  729. 

What  is  the  root  of  1849  ? 

• 

18,49  (40  +  3  = 

43  root. 

16,00 

249  (80  divisor. 
43  X  43  =  1849. 

In  this  example,  the  second  power  of  the  tens  will  be  found 
in  the  1800.  30  X  30  =  900  ;  40  X  40  =  1600  ;  50  X  50  =: 
2500.  The  greatest  second  power  in  1800  is  1600,  the  root  of 
which  is  40.  Write  40  in  the  place  of  a  quotient.  Subtract 
1600  from  1829.  The  remainder  is  249,  which  divided  by 
twice  40,  or  80,  gives  3.  Add  3  to  the  root,  and  raise  the 
whole  to  the  second  power.  43  X  43  ==  1849.  Therefore  43 
is  the  root  required. 

It  is  evident  that  the  result  will  not  be  affected,  if  instead  of 
writing  40  in  the  root  at  first,  we  omit  the  zero,  and  then  sub- 
tract the  second  power  of  4,  viz.  16  from  the  18,  omitting  the 
two  zeros  which  come  under  the  other  period.  Then  to  form 
the  divisor,  the  4  may  be  doubled,  and  the  divisor  will  be  8  in- 
stead of  80,  and  the  dividend  must  be  24,  the  right  hand  figure 
being  rejected. 


136 


Algebra. 


XXVIII. 


Operation, 

18,49  (43  root. 
16 


Ans,  38. 


Dividend  =    24,9  (8  divisor, 
i  43  X  43=z     18  49. 

Examples, 

1.  What  is  the  root  of  1444  .^ 

2.  What  is  the  root  of  7396  ^ 

3.  What  is  the  root  of  361  ? 

4.  What  is  the  root  of  3249  ? 

5.  What  is  the  root  of  7921  .? 

6.  What  is  the  root  of  8281  ? 

The  second  power  of  a  ~\- b  -^  c,  oi  {a  -\~  b  -^^  c)  (a -{-  b -^  c) 

a' ■j-2ab-{-b' -{-2ac-j'2bc  +  c'  — 

a'  -^2ab-\-b'  +2{a-^b)c^c\ 

To  find  the  second  power  of  726 

Let  a  =  700,     b  =  20,  and  c  =  6. 


IS 


a^  =  700  X  700 
2  a  6  =  2  X  700  X  20 
b'  =z  20  X  20 

2  (a  -f-  6)  c  =  2  X  (700  -f  20)  X  6 
c*  =  6X6 


726 
726 

4356 
1452 
5082 


=  490000 
=  28000 
=r  400 
=  8640 
=    36 


527076 


527076 


m 


';0: 


XXVIII.  Extraction  of  the  Second  Root.  137 

The  first  three  terms  of  the  formula,  viz. 
a'  +  2  «  6  +  5^ 

are  the  second  power  of  a  -|-  ^  or  of  the  hundreds  and  tens, 
viz.  720.  The  second  power  of  720  can  have  no  significant 
figure  below  hundreds,  and  the  significant  figures  of  the  se- 
cond power  of  720  and  of  72  are  the  same ;  the  former  is 
518400,  the  latter  5184.  If  from  the  whole  number  527076 
the  two  right  hand  figures  be  rejected,  the  number  is  5270. 
This  contains  the  second  power  of  72  and  something  more,  viz. 
a  part  of  the  product  2  X  (700  +  20)  X  6  =  2  {a  -f  b)  c. 

The  method  of  procedure  then,  is  to  find  the  largest  root 
contained  in  5270.  The  first  three  terms  of  the  above  formula, 
viz.  «^  +  2  a  6  -f-  b^j  show,  that  this  is  to  be  found  by  the  me- 
thod given  above  for  finding  a  root  consisting  of  two  figures. 

62,70  (72 
49 


37,0  (14 
72  X  72  =  51,84 


86 

The  root  is  72,  and  the  remainder  is  86.  Annex  to  this  the 
two  figures  rejected  above,  and  it  becomes  8676.  This  con- 
tains 2  {a -{■  h)  c  -\- c^  ;  that  is, 

2  X  720  X  c  -f-  c'. 

If  8676  be  divided  by  2  X  720  =  1440,  the  quotient  will  be 
either  c  or  a  number  larger  by  1  or  2.  The  zero  on  the  right 
of  1440,  and  the  right  hand  figure  in  the  dividend  may  be 
omitted  without  affecting  the  quotient.  The  quotient  is  6. 
Put  6  into  the  root  and  raise  the  whole  to  the  second  power. 

726  X  726  =  527076 
12* 


^ 

¥■ 


rSS  Algebra.  XXVIII. 

Operation. 

52,70,76  (726  —  root. 
49 


1st.  dividend        37,0  (14  r=  1st  divisor. 

72  X  72  =  51,84 


2d  dividend  rz:     867,6  (144  r=  2d  divisor. 
726  X  726:=  527,076. 

There  is,  however,  a  method,  which  will  save  considerable 
labour  in  multiplying. 

In  the  last  example,  for  instance,  having  found  the  second 
figure  of  the  root  2,  instead  of  raising  the  whole  72  to  the  se- 
cond power,  we  may  abridge  it  very  much  by  observing,  that 
the  second  power  of  the  70,  answering  to  a^  in  the  formula,  has 
already  been  found  and  subtracted  ;  therefore  it  only  remains 
to  find  2ab  -}-  h~,  and  subtract  it  also.  But  the  140  is  2  a,  and 
the  figure  2  found  for  the  root  answers  to  b ;  therefore  if  we 
add  2  to  140,  it  becomes  142  z:z2a  -\-h.  If  this  be  now  mul- 
tiplied by  2  or  6,  it  becomes 

2  X  142  =284=:2a6-f  6^ 

This  completes  the  second  power  of  72,  which,  subtracted 
from  370,  leaves  86  as  before. 

Prepare  as  before,  and  find  the  third  figure  of  the  root.  Ob- 
serve that  the  2d  power  of  720  or  a^  -j-  ^  ^  ^  +  ^^  ^^^  already 
been  found  and  subtracted  ;  it  only  remains  to  find  the  other 
parts,  viz.  2  {a  -\' b)  c  -{■  r.  The  divisor  1440  answers  to  2 
(a  -f-  6).  Add  6,  the  figure  of  the  root  just  found,  to  this,  and 
it  becomes  1446,  answering  to  2  (a  -\- b)  -\-  c.  If  this  be  mul- 
tiplied by  6,  it  becomes  1446  X  6  =  8676  =:  2  (a  +  6)  c  -f-  r . 
This  completes  the  second  power  of  726,  which,  subtracted 
from  8676,  the  number  remaining  in  the  work,  leaves  nothing. 


XXVIII.  Extraction  of  the  Second  Root.  139 

Operation. 

52,70,76,  (726  root. 
49 


1st  dividend       370  14     1st  divisor. 

284  142     1st  multiplicand. 


2d  dividend         8676        144     2d  divisor 

8676     1446     2d  multiplicand. 


00 
The  same  principle  will  apply  when  the  root  consists  of  any 
number  of  figures  whatever.  ^ 

What  is  the  root  of  533837732164  : 

In  the  first  place  I  observe  that  the  second  power  of  the  tens 
can  have  no  significant  figure  below  hundreds,  therefore  the 
two  right  hand  figures  may  be  rejected  for  the  present.  Also 
the  second  power  of  the  hundreds  can  have  no  significant  figure 
below  tens  of  thousands,  therefore  the  next  two  may  be  reject- 
ed. For  a  similar  reason  the  next  two  may  be  rejected.  In 
this  manner  they  may  all  be  rejected  two  by  two  until  only  one 
or  two  remain.  Begin  by  finding  the  root  of  these,  and  pro- 
ceed as  above. 

Operation. 

53,38,37,73,21,64  (730642 
49 

43,8  (143 
42  9 

93,7  (1460 

9377,3  (14606 
8763  6 

613  72,1  (146124 
584  49  6 


29  22  564  (1461282 
29  22  564. 


140  Algebra.  XXVIII. 

After  separating  the  figures  two  by  two,  as  explained  above^ 
I  find  the  greatest  second  power  in  the  left  hand  division.  It 
is  49,  the  root  of  which  is  7.  I  subtract  49  from  53,  and 
bring  down  the  next  two  figures,  which  makes  438.  Now 
considering  the  7  as  tens,  I  proceed  as  if  I  were  finding  the 
root  of  5338':  that  is,  I  double  the  7,  which  makes  14  for  a 
.  divisor,  and  see  how  many  times  it  is  contained  in  43,  rejecting 
the  8  on  the  right.  I  find  3  times.  1  write  3  in  the  root  at  the 
right  of  7,  and  also  at  the  right  of  14.  I  multiply  143  by  3, 
and  subtract  the  product  from  438.  I  then  bring  down  the 
next  two  figures,  which  make  937.  I  double  73,  or,  which  is 
the  same  thing,  I  double  the  3  in  143 ;  for  the  7  was  doubled 
to  find  14.  This  gives  146  for  a  divisor.  I  seek  how  many 
times  146  is  contained  in  93,  rejecting  the  7  on  the  right,  as 
before.  I  find  it  is  not  contained  at  all.  I  write  zero  in  the 
root,  and  also  at  the  right  of  146.  I  then  bring  down  the  next 
two  figures.  I  seek  how  many  times  1460  is  contained  in  9377, 
rejecting  the  3  on  the  right.  I  find  6  times.  I  write  6  in  the 
root,  and  at  the  right  of  1460,  and  multiply  14606  by  6,  and 
subtract  the  product  from  93773.  I  then  bring  down  the  next 
two  figures,  and  double  the  right  hand  figure  of  the  last  multi- 
plicand, and  proceed  as  before  ;  and  so  on,  till  all  the  figures 
are  brought  down.  The  doubling  of  the  right  hand  figure  of 
the  last  multiplicand,  is  always  equivalent  to  doubling  the  root 
as  far  as  it  is  found. 

From  the  above  examples,  we  derive  the  following  rule  for 
extracting  the  second  root. 

1st.  Beginning  at  the  right,  separate  the  number  into  parts  of 
two  figures  each.  The  left  hand  part  may  consist  of  one  or  two 
figures. 

2nd.  Find  the  greatest  second  power  in  the  left  hand  party  and 
write  its  root  as  a  quotient  in  division.  Subtract  the  second  power 
from  the  left  hand  part. 

3d.  Bring  down  the  two  next  figures  at  the  right  of  the  remain- 
der. DouhJe  the  root  already  found  for  a  divisor.  See  how  many 
times  the  divisor  is  contained  m  the  dividend  rejecting  the  right  Jiand 
figure.  Write  the  result  in  the  root,  at  the  right  of  the  figure  pre- 
viously found,  and  also  at  the  right  of  the  divisor. 

4th.  Multiply  the  divisor,  thus  augmented,  by  the  last  figure 
of  the  root,  and  subtract  the  product  from  the  whole  dividend. 


XXVIIL  Extraction  of  the  Second  Root.  141 

5th.  Bring  down  the  next  two  figures  as  hefore^  to  form  a  new 
dividend,  and  doMe  the  root  already  found,  for  a  divisor,  and  pro- 
ceed as  before.  The  root  will  be  doubled,  if  the  right  hand  figure 
of  the  last  divisor  be  doubled. 

If  it  happens  that  the  divisor  is  not  contained  in  the  dividend 
when  the  right  hand  figure  is  rejected,  a  zero  must  be  written 
in  the  root,  and  also  at  the  right  of  the  divisor ;  and  the  next 
figures  must  be  brought  down,  and  then  a  new  trial  made. 

If  it  happens  that  the  figure  annexed  to  the  root  is  too  small, 
it  may  be  discovered  as  follows. 

The  second  power  oi a  -\-  Wsa"  -\-2  a  -{•  I, 

That  is,  if  we  have  the  second  power  of  any  number,  the  se- 
cond power  of  a  number  larger  by  1,  is  found  by  multiplying 
the  first  number  by  2,  increasing  the  product  by  I,  and  adding 
it  to  the  power.  For  example,  the  second  power  of  10  is  100  ; 
the  second  power  of  11  is  100  +  2  x  10  +  1  =  121.  The 
second  power  of  12  is  121  -f  2  X  11  -f-  1  =  144,  &c. 

If  then  the  remainder,  after  subtraction,  is  equal  to  twice  the 
root  already  found  plus  1 ,  or  greater,  the  last  figure  of  the  root 
must  be  increased  by  1. 

In  the  last  example,  the  first  dividend  was  43,8  and  the  di- 
visor 14  ;  the  figure  put  in  the  root  was  3,  and  the  remainder 
was  9.  If  2  instead  of  3  had  been  put  in  the  root,  the  remain- 
der would  have  been  154,  which  is  considerably  larger  than 
twice  72,  and  would  have  shown,  that  the  figure  should  be  3 
instead  of  2. 

There  are  many  numbers,  of  which  the  root  cannot  be  ex- 
actly assigned  in  whole  or  mixed  numbers.  Thus  2,  3,  5,  6, 
7,  have  no  assignable  roots.  That  is,  no  number  can  be  found, 
which,  multiplied  into  itself,  shall  produce  either  of  these  num- 
bers. This  is  the  case  with  all  whole  numbers,  which  have 
not  an  exact  root  in  whole  numbers. 

This  may  be  proved,  but  the  demonstration  is  so  difficult, 
that  few  learners  would  comprehend  it  at  this  stage  of  their 
progress.  The  proof  may  be  found  in  Lacroix's  Algebra. 
The  learner,  however,  may  easily  satisfy  himself  by  trial.  We 
shall  soon  find  a  method  of  approximating  the  roots  of  these 
numbers,  sufficiently  near  for  all  purposes. 


145  ^    Algebra.  XXiX. 

XXIX.     Extraction  of  the  second  Root  of  Fractions. 

Fractions  are  multiplied  together  by  multiplying  their  nu- 
merators together,  and  their  denominators  together.  Hence 
the  second  power  of  a  fraction  is  found  by  multiplying  the  nu- 
merator into  itself,  and  the  denominator  into  itself;  thus  the 

second  power  of  |  is  |  x  |  =  2 s*     The  second  power  of 

__  is  ^  X  ~  =  ^-  Hence  the  root  of  a  fraction  is  found 
o       0         h         0^ 

by  extracting  the  root  of  the  numerator,  and  of  the  denomina- 
tor ;  thus  the  root  of  |^  is  f . 

If  either  the  numerator  or  denominator  has  no  exact  root, 
the  root  of  the  fraction  cannot  be  found  exactly.  Thus  the 
root  of  If  is  between  f  and  |  or  1.     It  is  nearest  to  |. 

The  denominator  of  a  fraction  may  always  be  rendered  a 
perfect  second  power,  so  that  its  root  may  be  found  ;  and  for 
the  numerator,  the  number  which  is  nearest  to  the  root  must 
be  taken.  Suppose  it  is  required  to  find  the  root  of  |.  If 
both  terms  of  the  fraction  be  multiplied  by  5,  the  value  of  thet 
fraction  will  not  be  altered,  and  the  denominator  will  be  a  per- 
fect second  power, 


3   —  Lk 

6     2  5' 


The  root  is  nearest  |.     This  is  exact,  within  less  than  i. 

If  it  is  necessary  to  have  the  root  more  exactly  ;  after  the 
fraction  has  been  prepared  by  multiplying  both  its  terms  by 
the  denominator,  we  may  again  multiply  both  its  terms  by  some 
number  that  is  a  perfect  second  power.  The  larger  this  num- 
ber, the  more  exact  the  result  will  generally  be. 

3  —  11 

5     2  5* 

If  both  terms  be  multiplied  by  144,  which  is  the  second 
power  of  12,  it  becomes  ||^|,  the  root  of  which  is  nearest  to 
^^.     This  is  the  true  root  within  less  than  ^V- 

We  may  approximate  in  this  way  the  roots  of  whole  num- 
bers, whose  roots  cannot  be  exactly  assigned. 

If  it  is  required  to  find  the  root  of  2,  we  may  change  it  to  a 
fraction,  whose  denominator  is  a  perfect  second  power. 

Q  _   2  8  8 
^  —  T4T- 


XXIX.         Extraction  of  the  Second  Root  of  Fractions.  143 

The  root  of  f||  is  nearest  to  i|  ==  IfV-  This  differs  from 
the  true  root  by  a  quantity  less  than  j'^.  If  greater  exactness 
is  required,  a  number  larger  than  144  may  be  used. 

1.  What  is  the  root  of  j\\  ?  Ans.  j\, 

2.  What  is  the  root  of  iff  ? 

3.  What  is  the  root  of  13i|f  rr  VoV  - 

4.  What  is  the  root  of  28|ff  ? 

5.  What  is  the  approximate  root  of  f  ? 

6.  What  is  the  approximate  root  of  f  |  ? 

7.  What  is  the  approximate  root  of  3f  ? 

8.  What  is  the  approximate  root  of  ITy^-  ? 

9.  What  is  the  approximate  root  of  3  ? 

10.  What  is  the  approximate  root  of  7  ? 

11.  What  is  the  approximate  root  of  417  .^ 

The  most  convenient  numbers  to  multiply  by,  in  order  to 
approximate  the  root  more  nearly,  are  the  second  powers  of 
10,  100,  1000,  .&c.,  which  are  100,  10000,  1000000,  &c.  By 
this  means,  the  results  will  be  in  decimals. 

To  find  the  root  of  2  for  instance,  first  reduce  it  to  hun- 
dredths. 

2  =z  f.^^,  the  approximate  root  of  which  is  |f  =:  1.4. 

Again  2  =:  f^||f,  the  approximate  root  of  which  is  iff  = 
1.41. 

Again,  2  =  f  f  f  f  f  f  f ,  the  approximate  root  of  which  is  |f  f  f 
=  1.414. 

In  this  way  we  may  approximate  the  root  with  sufficient  ac- 
curacy for  every  purpose.  But  we  may  observe,  that  at  every 
approximation,  two  more  zeros  are  annexed  to  the  number.  In 
fact,  if  one  zero  is  annexed  to  the  root,  there  must  be  two  an- 
nexed to  its  power  ;  for  the  second  power  of  10  is  100,  that 
of  100  is  10000,  &c. 

This  enables  us  to  approximate  the  root  by  dtjcimals,  and 
we  may  annex  the  zeros  as  we  proceed  in  the  work,  always  an- 
nexing two  zeros  for  each  new  figure  to  be  found  in  the  root,  in 
the  same  manner  as  two  figures  are  brought  down  in  whole 
numbers. 

■  r 

Trie. 

UNIVERSITY 

OF 


144  .ilgebra.  .       XXIX. 

The  root  of  2  then  may  be  found  as  follows. 
2  (1.41421,  &c.  root. 


10,0  (24 
96 


40,0  (281 
28  1 

11  90,0(2824 
11  29  6 

60  40,0  (28282 
56  56  4 


3  83  60,0  (282841 
2  82  84  1 


1  00  75  9 

12.  What  is  the  approximate  root  of  28  ? 

1 3.  What  is  the  approximate  root  of  243  ? 

14.  What  is  the  approximate  r*ot  of  27068  ? 

15.  What  is  the  approximate  root  of  243|  ? 

C)A<^3    94."^      375       243375.0    243375000      ^  f. 

X>^Oj    ^^U    T0'0"0     lOUOO         1000000      >    »-»''^' 

The  approximate  root  of  which  is  VVoV  =  15.6,  &.c. 

But  it  is  plain  that  this  may  be  performed  in  the  same  man- 
ner as  the  above.  For  if  the  number  243375000  be  prepared 
in  the  usual  way,  it  stands  thus  ;  2,43,37,50,00.     Now 

nffUU"  =243.375000. 

If  we  take  this  number  and  begin  at  the  units  and  point  to- 
wards the  left,  and  then  towards  the  right  in  the  same  man- 
ner, the  number  will  be  separated  into  the  same  parts,  viz. 
2,43.37,50,00.  The  root  of  this  number  may  be  extracted  in 
the  usual  way,  and  continued  to  any  number  of  decimal  places 
by  annexing  zeros. 


XXX.  Questions  jproducing  Pure  Eqtuitions,  145 

N.  B.  The  decimal  point  must  be  placed  in  the  root,  be- 
fore the  first  two  decimals  are  used.  Or  the  root  must  con- 
tain one  half  as  many  decimal  places  as  the  power,  counting 
the  zeros  which  are  annexed. 

16.  What  is  the  approximate  root  of  213.53  ? 

17.  What  is  the  approximate  root  of  726f  ? 

18.  What  is  the  approximate  root  of  17y\  f 

19.  What  is  the  approximate  root  of  3|i  ? 

20.  What  is  the  approximate  root  of  |  ? 

21.  What  is  the  approximate  root  of  f  ? 

22.  What  is  the  approximate  root  of  jji ' 

23.  What  is  the  approximate  root  of  yy^a  ^ 

XXX.     Questions  producing  pure  Equations  of  the  Second 

Degree, 

1.  A  mercer  bought  a  piece  of  silk  for  £16.  4s. ;  and  the 
number  of  shillings  which  he  paid  per  yard,  was  to  the  number 
of  yards,  as  4  to  9.  How  many  yards  did  he  buy,  and  what 
was  the  price  of  a  yard  ? 

Let  X  =  the  number  of  shillings  he  paid  per  yard. 

9  X 
Then   —  =  the  number  of  yards. 

9  a;* 
The  price  of  the  whole  will  be =  324  shillings 

x'  =  144 
a:  ^    12 

if.  =    27. 
4 

Am,  27  yards,  at  12s.  per  yard. 

2.  A  detachment  of  an  army  was  marching  in  regular  co- 
lumn, with  5  men  more  in  depth  than  in  front ;  but  upon  the 
enemy  coming  in  sight,  the  front  was  increased  by  845  men ; 
and  by  this  movement  the  detachment  was  drawn  up  in  5  lines. 
Required  the  number  of  men. 

13 


14^  .  Algebra,  XXX. 

Iiiet  a;  =  the  number  in  front ; 
'  then  X  -^  5  =  the  number  in  depth ; 
j7*  -j-  5  0?  =  the  whole  number  of  men. 
Again  x  -j-  845  =  the  number  in  front  after  the  movement ; 
And  5  a?  -f  4225  =  the  whole  number.     }i 

x''  +  5x=z5x-\-  4225 
x^    =4225 
a?     =      65 

The  number  of  men  =  5  a;  +  4225  =  4550. 

3.  A  piece  of  land  containing  160  square  rods,  is  called  an 
acre  of  land.  If  it  were  square,  what  would  be  the  length  of 
one  of  its  sides  ? 

Let  X  =  one  side. 

a?*=l60      " 
X  =z  12649  -f 

Ans,  The  side  is  12.649  +  rods.  It  cannot  be  found  exactly, 
because  160  is  not  an  exact  2d  power. 

This  is  exact  within  less  than  yoVo  ^^  ^  ^^^'  It  might  be 
carried  to  a  greater  degree  of  exactness  if  necessary. 

4.  What  is  the  side  of  a  square  field,  contaming  17  acres  ? 

5.  There  is  a  field  144  rods  long  and  81  rods  wide  ;  what 
would  be  the  side  of  a  square  field,  whose  content  is  the  same  ? 

6.  A  man  wishes  to  make  a  cistern  that  shall  contain  100 
gallons,  or  23100  cubic  inches,  the  bottom  of  which  shall  be 
square,  and  the  height  3  feet.  What  must  be  the  length  of 
one  side  of  the  bottom  ? 

f.  A  certain  sum  of  money  was  divided  every  week  among 
the  resident  members  of  a  corporation.  It  happened  one  week 
that  the  number  resident  was  the  root  of  the  number  of  dollars 
to  be  divided.  Two  men  however  coming  into  residence  the 
week  after,  diminished  the  dividend  of  each  of  the  former  indi- 
viduals 1  i  dollars.     What  was  the  sum  to  be  divided  ? 

Let  X  =  the  number  of  dollars  to  be  divided ; 


XXX.  Questions  producing  Pure  Equations.  147 

then  x^  =  the  number  of  men  resident,  aad  also  the  sum 
each  received. 

The  root  of  a?  is  properly  expressed  by  the  fractional  index 
J.  For  it  has  been  observed,  that  when  the  same  letter  is 
found  in  two  quantities  which  are  to  be  multiplied  together, 
the  multiplication  is  perforrped,  as  respects  that  letter,  by 
adding  the  exponents.      Thus  a  X  a  z=z  a}  +  ^  =.a^  *,  x^  X  a?' 

—  a?*  +  ^  =:  a:*,  &;c.     Applying  the  same  rule  ;  if  x^  represents 

a  root  or  first  power,  the  second  power  or  x^  X  a?    =:  a?^  '  ^ 

=  0?*  or  X. 

The  second  power  of  a  letter  is  formed  from  the  first  by 
multiplying  its  exponent  by  2,  because  that  is  the  same  as 
adding  the  exponent  to  itself.  Thus  a^  x  a^  =  a^  + '  =  a'  ^  ' 
=  a'.  This  furnishes  us  with  a  simple  rule  to  find  the  root  of 
a  literal  quantity ;   which  is,  to  divide  its  exponent  by  2. 

2  4. 

Thus  the  root  of  a*  is  a^  zr:  a' ;  the  root  of  a*  =z  a^  :=±a^',  the 

root  of  aMs  a^  =  a^  &c.     By  the  same  rule,  the  root  of  a}  is 

a^  ;   the  root  of  aMs  a^  ;  the  root  of  a**  is  a^  ;  the  root  of  a 

■» 
is  a2,  &c. 

In  the  above  example 
X  =  the  number  of  dollars  to  be  divided  ; 

and  x^  =  the  number  of  men  resident ; 

—    i    i        X 
^~  ^    X  __  -jn2  _.  ijjg  number  of  dollars  each  received. 

a?*  +  2  =  the  number  of  men  the  succeeding  week  ; 

=  the  number  of  dollars  each  received  the  latter  week; 

^*+2 

Hence  by  the  conditions 

i        4  X 


3        i  +  2 


X' 


148  Algehra.  XXX. 

3     ^  3 

3^3 

-1^  +  2.4-1  =  0 
3    ^  3 

-1^^2.4  =  1 

3         ^  3 

-^  4  a:*  +  6  X*  =  8 


*   - 


2  a?' 


a?*.X  ^*  =  a?*  +  *  =  a?  =  4  X  4  =  16. 

./?n5.  $16. 

Instead  of  making  x  =  the  number  of  dollars,  we  might 
make, 

x^  =  the  number  of  dollars  ; 
then  X  =z  the  number  of  men  resident,  &c. 
Then  we  have 

4  a^* 


X  — 

'  3 

x  + 

2 

a^- 

4  a? 
'  3 

+  2^7- 

__  8 
3 

=  a^ 

x»- 

_4a? 
"  3 

+  2a?- 

2a: 
a? 
a:« 

_  8 
3 

=  8 

=  4 

=  16. 

$16, 

as  before. 

XXX.  Questions  producing  Pure  Equations,  149 

8.  Two  men,  A  and  B,  lay  out  some  money  on  speculation. 
A  disposes  of  his  bargain  for  £11,  and  gains  as  much  per  cent 
as  B  lays  out ;  B's  gain  is  £36,  and  it  appears  that  A  gains 
four  times  as  much  per  cent,  as  B.  Required  the  capital  of 
each. 

9.  There  is  a  rectangular  field  containing  3G0  square  rods, 
and  whose  length  is  to  its  breadth  as  8  to  5.  Required  the 
length  and  breadth. 

10.  There  are  two  square  fields,  the  larger  of  which  contains 
13941  square  rods  more  than  the  smaller,  and  the  proportion 
of  their  sides  is  as  15  to  8.     Required  the  sides. 

11.  There  is  a  rectangular  room,  the  sum  of  whose  length 
and  breadth  is  to  their  difference  as  8  to  1  ;  if  the  room  were 
a  square  whose  side  is  equal  to  the  length,  it  would  contain 
128  square  feet  more  than  it  would,  if  it  were  only  equal  to  the 
breadth.     Required  the  length  and  breadth  of  the  room. 

12.  There  is  a  rectangular  field,  whose  length  is  to  its 
breadth  in  the  proportion  of  6  to  5.  A  part  of  this,  equal  to  ^ 
of  the  whole,  being  planted,  there  remain  for  ploughing  625 
square  yards.     What  are  the  dimensions  of  the  field  f 

13.  A  charitable  person  distributed  a  certain  sum  amongst 
some  poor  men  and  women,  the  number  of  whom  were  in  the 
proportion  of  4  to  5.  Each  man  received  one  third  as  many 
shillings  as  there  were  persons  relieved  ;  and  each  woman  re- 
ceived twice  as  many  shillings  as  there  were  v/omen  more  than 
men.  The  men  received  all  together  18s.  more  than  the  wo- 
men.    How  many  were  there  of  each  ? 

14.  A  man  purchased  a  field  whose  length  was  to  the 
breadth  as  8  to  5.  The  number  of  dollars  paid  per  acre  was 
equal  to  the  number  of  rods  in  the  length  of  the  field  ;  and  the 
number  of  dollars  given  for  the  whole,  was  equal  to  13  times 
the  number  of  rods  round  the  field.  Required  the  length  and 
breadth  of  the  field. 

15.  There  is  a  stack  of  hay  whose  length  is  to  its  breadth  as 
5  to  4,  and  whose  height  is  to  its  breadth  as  7  to  8.  It  is  worth 
as  many  cents  per  cubic  foot  as  it  is  feet  in  breadth  ;  and  the 
whole  is  worth,  at  that  rate,  224  times  as  many  cents  as  there 
are  square  feet  on  the  bottom.  Required  the  dimensions  of 
the  stack. 

13* 


150  Algebra.  XXXi. 

16.  There  is  a  field  containing  108  square  rods,  and  the 
sum  of  the  length  and  breadth  is  equal  to  twice  the  difference. 
Required  the  length  and  breadth. 

17.  There  are  two  numbers  whose  product  is  144,  and  the 
quotient  of  the  greater  by  the  less  is  16.  What  are  the  num- 
bers ? 

XXXI.     Questions  producing  Pure  Equations  of  the  Third 
Degree. 

1 .  A  number  of  boys  set  out  to  rob  an  orchard,  each  carry- 
ing as  many  bags  as  there  were  boys  in  all,  and  each  bag  ca- 
pable of  containing  8  times  as  many  apples  as  there  were  boys. 
They  filled  their  bags,  and  found  the  whole  number  of  apples 
Was  1000.     How  many  boys  were  there  ? 

Let  X  =:  the  number  of  boys  ; 
then  X  X  x=za!^  z=z  the  number  of  bags ; 
and  8  >r  X  a?'  =  8  «^  =  the  number  of  apples. 
By  the  conditions 

8^^=1000 
a?'=    125 
or  XXX  z:z    125. 

In  this  equation,  the  unknown  quantity  is  raised  to  the  third 
power;  and  on  this  account  is  called  an  equation  of  the  third 
degree. 

In  order  to  find  the  value  of  a:  in  this  equation,  it  is  necessa- 
ry to  find  what  number  multiplied  twice  by  itself  will  make  125. 
By  a  few  trials  we  find  that  5  is  the  number  ;  for 

5  X  5  X  5  =r  125 

therefore  ^  =  5.  Ms.  5  boys. 

2.  Some  gentlemen  made  an  excursion  ;  and  every  one  took 
the  same  sum  of  money.  Each  gentleman  had  as  many  ser- 
vants attending  him  as  there  were  gentlemen ;  and  the  nuni- 
ber  of  dollars  which  each  had,  was  double  the  number  of  all 


XXXI.  Extraction  of  the  Third  Root,  15# 

the  servants  ;  and  the  whole  sum  of  money  taken  out  was 
$1458.     How  many  gentlemen  were  there  ? 

Am.  9  gentlemen. 

3.  A  poulterer  bought  a  certain  number  of  fowls.  The  first 
year  each  fowl  had  a  number  of  chickens  equal  to  the  original 
number  of  fowls.  He  then  sold  the  old  ones.  The  next  year 
each  of  the  young  ones  had  a  number  of  chickens  equal  to 
once  and  one  half  the  number  which  he  first  bought.  The 
whole  number  of  chickens  the  second  year  was  768.  What 
was  the  number  of  fowls  purchased  at  first  ? 

It  appears  that  in  equations  of  the  third  degree,  as  in  those 
of  the  second  degree,  the  power  of  the  unknown  quantity  must 
first  be  separated  from  the  known  quantities,  and  made  to  stand 
alone  in  one  member  of  the  equation,  by  the  same  rules  as  the 
unknown  quantity  itself  is  separated  in  simple  equations. 
When  this  is  done,  the  first  power  or  the  root  must  be  found, 
and  the  work  is  finished. 

Extraction  of  the  Third  Root, 

The  third  power  of  a  quantity  is  easily  found  by  multiplica- 
tion, but  to  return  from  the  power  to  the  root,  is  not  so  easy. 
It  must  be  done  by  trial,  in  a  manner  analogous  to  that  em- 
ployed for  the  root  of  the  second  power. 

We  shall  hereafter  have  occasion  to  speak  of  the  root  of  the 
fourth  power,  of  the  fifth  power,  &c.  In  order  to  distinguish 
them  the  more  readily,  we  shall  call  the  root  of  the  second 
power,  the  second  root  of  the  quantity  ;  that  of  the  third  power, 
the  third  root,  that  of  the  fourth  power,  the  fourth  root,  &,c.  To 
preserve  the  analogy,  we  shall  sometimes  call  the  root  of  the 
first  power,  the  first  root. 

N.  B.  The  first  power,  and  the  first  root,  are  the  same 
thing,  and  the  same  as  the  quantity  itself. 

It  always  has  been,  and  is  still  the  practice  of  mathemati- 
cians, to  call  the  second  root  the  square  root,  and  the  third 
root  the  cube  root,  and  sometimes,  though  not  so  universally, 
the  fourth  root  the  hi-quadrate  root.  But  as  these  terms  are 
unappropriate,  they  will  not  be  used  in  this  treatise. 

When  the  root  consists  of  but  one  figure,  it  must  be  found 
by  trial.     When  the  root  consists  of  more  than  one  place,  it 


162  Algehra.  XXXI. 

must  still  be  found  by  trial,  but  rules  may  be  made,  which  will 
reduce  the  number  of  trials  to  very  few,  as  has  been  done 
above  ibr  the  second  root. 

In  order  to  find  the  rules  for  extracting  the  third  root,  it  will 
be  necessary  to  observe  how  the  third  power  is  formed  from  the 
first,  when  the  first  consists  of  several  figures. 

Let  a  =  30  and  6  =  5;  then  a  -|-  6  =  35. 

(a  -I-  6)'  =  a'  +  3  a'  6  +  3  a  &»  -f  b\     Art.  XIII. 

a^  =  30  X  30  X  30  =  27000 

3a'6=3x30x30x5     =  13500 

3«6»=    3X30X    5x5     =    2250 

6'  =  5X5X5=      125 


42875 


Hence  it  appears,  that  the  third  power  of  a  number  consist- 
ing of  units  and  tens,  contains  the  third  power  of  the  tens, 
plus  three  times  the  second  power  of  the  tens  multiplied  by  the 
units,  plus  three  times  the  tens  multiplied  by  the  second  power 
of  the  units,  plus  the  third  power  of  the  units. 

Farther,  the  third  power  of  10,  which  is  the  smallest  number 
with  two  places,  is  1000,  which  consists  of  four  places ;  and 
the  third  power  of  100,  is  1000000,  which  consists  of  seven 
places.  Hence  the  third  power  of  tens  will  never  be  less  than 
1000,  nor  so  much  as  1000000. 

If,  therefore,  there  are  tens  in  the  root,  their  power  will  not 
be  found  below  the  fourth  place ;  and  if  the  root  consists  of 
tens  without  units,  there  will  be  no  significant  figure  below 
1000. 

To  trace  back  again  the  number  42875,  the  root  of  the  tens 
will  be  found  in  the  42000,  and  this  must  be  found  by  trial. 

30  X  30  X  30  =  27000,  and  40  X  40  X  40  =  64000. 

The  largest  third  power  in  42000  is  27000,  the  root  of  which 
is  30.  Now  1  subtract  27000  from  42875,  and  the  remainder 
is  15875,  which  contains  the  product  of  three  times  the  second 
power  of  the  tens  by  the  units,  plus,  &c.  If  it  contained  ex- 
actly three  times  the  second  power  of  the  tens  multiplied  by 
the  units,  the  units  of  the  root  would  be  found  immediately  by 


XXXL  Extraction  of  the  Third  Root,  153 

dividing  this  remainder  by  three  times  the  second  power  of  the 
tens ;  for  3  a^  6  divided  by  3  a^  gives  h.  As  the  other  parts 
however  will  always  be  small  in  comparison  with  this,  if  we 
divide  the  remainder  by  three  times  the  second  power  of  the 
tens,  we  shall  be  able  to  judge  very  nearly  what  is  the  root, 
and  the  number  of  trials  will  be  limited  to  very  few. 

30  X  30  =  900,  and  900  X  3  =  2700  and  15875  divided  by 
2700  gives  5.  I  now  add  the  5  to  the  root  and  it  becomes  35. 
To  see  if  this  is  right,  I  raise  35  to  the  third  power.  35  X  35 
X  35  =  42875,  therefore  35  is  the  true  root. 

4.  What  is  the  third  root  of  79507  ? 

Operation, 

79,507  (40 -I- 3  =  43  root. 
64,000 


15,507  (40  X  40  X  3  =  4800  divisor. 
43  X  43  X  43  =  79,507. 

As  the  number  consists  of  five  places,  the  power  of  the  tens 
must  be  sought  in  the  79000. 

The  greatest  third  power  in  79000  is  64000,  the  root  of  which 
is  40.  I  subtract  64000  from  79507  and  there  remains  15507, 
which  I  divide  by  three  times  the  second  power  of  40,  viz. 
4800,  and  obtain  a  quotient  3,  which  I  add  to  40.  I  raise  43 
to  the  third  power,  and  find  that  it  gives  79507.  If  it  produced 
a  number  larger  or  smaller,  I  should  put  a  smaller  or  larger 
number  in  place  of  3  and  try  it  again. 

5.  What  is  the  third  root  of  35791 1 .? 

6.  What  is  the  third  root  of  5832  .? 

^  7.  What  is  the  third  root  of  941192  .? 
8.  What  is  the  third  root  of  34965783 .? 

It  was  observed  above,  that  the  third  power  of  10  is  1000 ; 
the  third  power  of  100  is  1000000  ;  that  of  1000  is  1000000000, 
(fee.  That  is,  the  third  power  of  a  number  consisting  of  one 
figure  cannot  exceed  three  places  ;  that  of  a  number  consist- 
ing of  two  places  cannot  contain  less  than  4  places  nor  more 


154  Algebra.  XXXL 

than  6  ;  that  of  3  places  cannot  contain  less  than  7  nor  more 
than  9  places,  &c. 

Hence  we  may  know  immediately  of  how  many  places  the 
third  root  of  any  given  number  will  consist,  by  beginning  at 
the  right  and  separating  the  number  into  parts  of  3  places 
each.     The  left  hand  part  will  not  always  contain  3  places. 

In  the  present  instance,  the  number  34,965,783,  thus  divided 
consists  of  three  parts,  therefore  the  root  will  contain  3  places 
or  figures. 

In  the  formula  {a  +  by  =  a^  -\- Sa'b -\-Sab' -\-  b\  if  we 
consider  a  as  representing  the  hundreds  of  the  root,  and  b  the^ 
tens  and  units,  we  observe  that  the  third  power  consists  of  the 
third  power  of  the  hundreds,  plus  3  times  the  second  power  of 
the  hundreds,  multiplied  by  the  u^its  and  tens,  &c. 

Hence  we  shall  find  the  hundreds  of  the  root  by  finding  the 
highest  third  power  contained  in  the  34,000,000,  and  taking  its 
root. 

The  largest  third  power  is  27,000,000,  the  root  of  which  is 
300.  Subtracting  27,000,000  from  the  whole  sum,  the  remain- 
der is  7,965,783.  If  this  contained  exactly  3  a^  b,  that  is,  3 
times  the  second  power  of  the  hundreds  by  the  tens  and  units, 
the  other  two  figures  of  the  root  might  be  found  immediately 
by  division.  As  it  is,  it  is  evident,  that  it  will  enable  us  to 
judge  very  nearly  what  the  next  figure,  or  tens,  of  the  root 
must  be,  and  its  correctness  must  be  proved  by  trial.  ' 

300  X  300  X  3  =  270000. 

7,965,783  divided  by  270000  gives  for  the  first  figure  of  the 
quotient  2,  which  being  the  tens  is  20.  This  added  to  the  root 
already  found  makes  320. 

If  in  the  above  formula,  we  consider  a  as  representing  the 
hundreds  and  tens  instead  of  the  hundreds  ;  and  b  as  repre- 
senting the  units  ;  it  shows  us  that  the  power  contains  the  third 
power  of  the  hundreds  and  tens,  plus  3  times  the  second  power 
of  the  hundreds  and  tens  multiplied  by  the  units,  &c.  In  the 
present  instance  a  =:  320.  If  now  we  subtract  the  third  power 
of  320  from  the  whole  sum,  viz.  34,965,783,  and  divide  the  re- 
mainder by  3  times  the  second  power  of  320,  we  shall  find  the 
other  figure,  or  units,  of  the  root.  When  we  have  raised  320 
to  the  third  power,  we  can  ascertain  whether  the  second  figure, 
2  is  right. 


XXXI,    .  Extradian  of  the  Third  Root.  156 

320  X  320  X  320  =  32768000. 

This  subtracted  from  34965783  leaves  2197783. 

320  X  320  X  3  =  307200. 

2197783  being  divided  by  307200  gives  a  quotient  7.  This 
added  to  320  gives  327  for  the  root. 

327  X  327  X  327  =  34,965,783. 

Therefore  the  result  is  correct. 

If  the  root  consists  q£  four  or  more  places,  the  same  mode 
of  reasoning  may  be  pursued  by  making  a  first  equal  to  the 
highest  figure  in  the  root,  and  b  equal  to  all  below,  until  the 
second  figure  of  the  root  is  obtained,  and  then  making  a  equal 
to  the  two  figures  already  obtained,  and  b  equal  to  the  rest, 
and  so  on. 

The  work  may  be  considerably  abridged  by  omitting  the 
zeros  in  the  work,  and  also  the  numbers  under  which  they  fall. 

The  work  of  the  above  example  will  stand  thus. 

Root. 
34,965,763  (300  +  20  +  7  =  327. 
—  27,000,000    3d  power  of  300 


1st  divid. 

7,965,783 

(270,000 

(1st  divisor     = 
-  i  300  X  300  X  3 

32,768,000 

3d  power  i 

(307,200 

of  320 

2d  divid. 

2,197,783 

(  2d  divisor     = 
i  320  X  320  X  3 

34,965,783  : 

=  3d  powei 

:  of  327. 

The  same  without  the  zeros. 

3d  power  of  3 

34,965,783 

27 

(327 

(27  1st  di^ 

(2072 

iBt  dividend         7,9 
3d  power  of  32  32  768 

irisor  =  3*  X  3 

2d  dividend 

2  197,7 
34,965,783 

<'  2d  divisor 
I  =  (32)*  X  3 

156  Algebra,  XXXI. 

As  the  third  power  can  have  no  significant  figure  below 
1000000,  and  as  the  third  power  of  300  and  3  have  the  same 
significant  figures,  I  raise  3  to  the  3d  power  and  subtract  it 
fi-om  34,  as  if  it  stood  alone.  Then,  to  form  the  divisor,  hun- 
dreds are  multiplied  by  hundreds,  therefore  there  can  be  no 
significant  figure  below  10000.  And  it  being  the  tens  of  the 
root  that  are  to  be  found,  it  is  sufficient  to  bring  down  one 
figure  of  the  next  period  to  form  the  dividend. 

Having  found  the  second  figure  of  the  root,  I  raise  32  to  the 
third  power,  and  subtract  it  from  34,965,  omitting  the  last  pe- 
riod, because  the  third  power  of  the  tens  can  have  no  signifi- 
cant figure  below  1000. 

To  form  the  second  divisor  I  multiply  the  second  power  of 
32  by  3.  For  the  dividend,  it  is  sufficient  to  bring  down  one 
figure  of  the  last  period  to  the  right  of  the  remainder,  because 
the  divisor,  being  tens,  multiplied  by  tens,  can  have  no  signifi- 
cant figure  below  100. 

JVote.  The  second  power  of  the  32  was  found  in  finding  its 
third  power. 

If  it  happens  that  the  divisor  is  not  contained  in  the  dividend, 
a  zero  must  be  put  in  the  root,  and  then  the  next  figure  must 
be  brought  down  to  form  the  dividend. 

Hence  we  obtain  the  following  rule  for  finding  the  third 
root. 

Prepare  the  number  by  beginning  at  the  right  and  separating  it 
into  parts  or  periods  of  three  figures  each,  putting  a  comma  or 
point  between.  The  left  hand  period  may  consist  of  one,  two,  or 
three  figures. 

Find  the  greatest  third  power  in  the  left  hand  period,  and  write 
the  root  in  the  place  of  a  quotient.  Subtract  the  power  from  the 
period.  To  the  remainder  bring  down  the  first  figure  of  the  next 
period  for  a  dividend.  Multiply  the  second  power  of  the  root 
already  found  by  three,  to  form  a  divisor.  See  how  many  times 
the  divisor  is  contained  in  the  dividend,  and  write  the  result  in  the 
root.  Raise  the  root,  thus  augmented,  to  the  third  power.  If  this 
is  greater  than  the  first  two  periods,  diminish  the  quotient  by  one  or 
more,  until  you  obtain  a  third  power,  which  may  be  subtracted  from 
the  first  two  periods.  Perform  the  subtraction,  and  to  the  right  of 
the  remainder  bring  down  tlie  first  figure  of  the  next  period  to 
form  a  dividend  and  divide  it  by  three  times  the  second  power  of 


XXXI.  Extraction  ofthe  Third  Root,  157 

the  two  figures  of  the  root,  and  write  the  quotient  in  the  root. 
Then  raise  the  whole  root  so  found,  to  the  third  power ;  and  if  it 
is  not  too  large,  subtract  it  from  the  first  three  periods  ;  if  it  is  too 
large,  diminish  the  root  as  before.  To  the  remainder  bring  down 
the  first  figure  of  the  fourth  period,  and  perform  the  same  series 
of  operations  as  before. 

If  at  any  time  it  should  happen  that  the  dividend,  prepared  as 

!  above,  does  not  contain  the  divisor,  a  zero  mu^t  be  placed  in  the 
root,  and  the  next  figure  brought  doWn  to  form  the  dividend- 

We  explained  a  method  in  the  extraction  of  the  second  root, 
\    more  expeditious  than  to  raise  the  root  to  the  second  power 
every  time  a  new  figure  is  obtained  in  the  root.     A  similar 
method  may  be  found  for  the  third  root,  though  it  is  rather  dif- 
ficult to  be  remembered. 

Let  «  =  30  and  J  =  7  ;  then 
{a  4-  by  =  (37)='  =  a^  ^  3  a'6  +  3  a&=  +  6'  =  50653 

To  find  the  third  root  of  50653,  find  the  first  figure  of  the 
root  as  explained  above.  Then  form  the  divisor  as  above,  and 
find  the  second  figure  of  the  root.  Then  instead  of  raising  the 
whole  to  the  third  power,  it  may  be  completed  from  the  work 
already  done.  The  third  power  of  the  first  figure  being  found 
and  subtracted,  the  remaining  part  is 

3a*6  +  3a6'-f-6'z=:5(3a'  +  3a6  +  b^ 

But  the  3  a^  has  already  been  found  for  the  divisor. 

\      We  must  now  find  3  a  6  and  b^ ;  add  all  together,  and  multi- 
ply the  sum  by  b,  and  the  third  power  will  be  completed. 

Operation. 

3  a=  =    3  X  30  X  30  =  2700     50,6  53  (30  +  7  =  37. 
3  a  6  =  30  X    7  X    3  =    630     27 


b^=    7  y    1  —     49     23  6,53  (2700  =  3  a% 

7  X  3379  z=i  23  6,53 

9.  What  is  the  third  root  of  34,965,783  ? 
14 


168 


Algebra, 


XXXI. 


We  have  seen  above,  that  when  the  root  is  to  consist  of  seve- 
fal  figures,  the  same  course  is  to  be  pursued  as  when  it  consists 
of  only  two. 


3a»  =  270000 
Sah=  18000 
6*     =        400 


Operation, 

34,965,783  (300  +  20  +  7  =  327, 
27 


288400 

20  =  6 


79,65 
57  68 


(2700  1st  divisor. 


5768000 


21  977,83    (307200  2d  divisor. 
21 977  83 


3(a'  +  2a6  +  5*)  = 
3a»  +  2x3a6  +  36' 

3  a*  =  270000 

2X  3a&=    36000 

3  ft'  =      1200 


2d  divisor    307200  =  3  X  320  X  320 

3  «''  =  307200 

3a6=      6720 

h*  =  49 


311969 
h=  7 

2197783 


Exampli 


es. 


10.  What  is  the  third  root  of  185193  ? 
U.  What  is  the  third  root  of  8365427  ? 
12.  What  is  the  third  root  of  77308776  ? 


XXXII.  Extraction  of  the  Third  Root.  15^ 

13.  What  is  the  third  root  of  1990865512  ? 

14.  What  is  the  third  root  of  513,345,176,343  f 

15    What  is  the  third  root  of  217,125,148,004,864  ? 

XXXII.  The  third  power  of  a  fraction  is  found  by  raising 
both  numerator  and  denominator  to  the  third  power.  Thus 
the  third  power  offis|XfX|=  xV^  • 

Hence  the  third  root  of  a  fraction  is  found  by  finding  the 
third  root  of  both  numerator  and  denominator..     The  third  of 


3V3=4. 


Examples. 

1.  What  is  the  third  root  of  fif  ? 

2.  What  is  the  third  root  of  ^f  !„  .^ 

3.  What  is  the  third  root  of  34f  J  =  VaV  ? 

4.  What  is  the  third  root  of  30f  i||f  ? 


It  was  remarked  with  regard  to  the  second  root  that,  when 
a  whole  number  has  not  an  exact  root  in  whole  numbers,  its 
root  cannot  be  exactly  found,  for  no  fractional  quantity  multi- 
plied by  itself  can  produce  a  whole  number.  The  same  is  true 
with  regard  to  all  roots,  and  for  the  same  reason. 

Hence  the  third  root  of  Y  cannot  be  found  exactly  because 
the  numerator  has  no  exact  third  root.  The  root  of  the  deno- 
minator is  2,  that  of  the  numerator  is  between  2  and  3,  nearest 
to  3.     The  approximate  root  is  f  or  1  J. 

6.  What  is  the  third  root  of  | .? 

In  this,  neither  the  numerator  nor  the  denominator  is  a  per- 
fect third  power  ;  but  the  denominator  may  be  rendered  a  per- 
fect third  power,  without  altering  the  value  of  the  fraction,  by 
multiplying  both  terms  of  the  fraction  by  49,  the  second  power 
of  the  denominator. 

3x  49  _  jm 
7  X  49  343  * 


160  Algehra.  XXXll. 

The  root  of  this  is  between  4  and  f,  nearest  to  the  former. 

It  is  evident  that  the  denominator  of  any  fraction  may  be 
rendered  a  perfect  third  power,  by  multiplying  both  its  terms 
by  the  second  power  of  the  denominator.  The  third  root  of 
a  whole  number  which  is  not  a  perfect  third  power,  may  be 
approximated  by  converting  the  number  into  a  fraction,  whose 
denominator  is  a  perfect  third  power. 

What  is  the  third  root  of  5  ? 

We  may  find  this  root  exact  within  less  than  y^  of  a  unit, 
by  converting  it  into  a  fraction,  whose  denominator  is  the  third 
power  of  12. 

(12)'  =  1728  5  =  ff  If 

The  root  of  f  f  ||  is  between  f^f  and  f  | ;  nearest  the  latter. 

The  most  convenient  numbers  to  multiply  by,  are  the  third 
powers  of  10,  100,  1000,  &c.  in  which  case,  the  fractional  part 
of  the  root  will  be  expressed  in  decimals,  in  the  same  manner 
as  was  shown  for  the  second  root.  The  multiplication  may  be 
performed  at  each  step  of  the  work.  For  each  decimal  to  be 
obtained  in  the  root,  three  zeros  must  be  annexed  to  the  num- 
ber, because  the  third  power  of  10  is  1000,  that  of  100, 
1000000,  &c. 

7.  The  third  root  of  5  will  be  found  by  this  method  as  fol- 
lows. 

5.000,000,000  (1.709  + 
3d  power  of  1  1 


S  X  (17)» 


1st  dividend  = 
3d  power  of  1.7 

40 
4.913 

(3     1st  divisor. 

2d  dividend  = 

3d      do. 

3d  power  1.709  = 

870 
8700 
:  4.991,443,829 

(867  2d  do.  = 
(867  3d  do. 

remainder 

.008  556  171, 

The  3d  root  of  5  is  1.709,  within  less  than  y^V(7  ^^^  "^lit. 
We  might  approximate  much  nearer  if  necessary.  The  other 
method  explained  in  the  last  article  may  be  used  if  preferred. 


XXXIII.  Extraction  of  the  Third  Root.  161 

8.  What  is  the  third  root  of  17J  ? 

The  fractional  part  of  this  number  must  first  be  changed  to 
a  decimal. 

171  =  17.75  =  VVoV  =  17.750. 

Hence  it  appears,  that  to  prepare  a  number  containing  deci- 
mals, it  is  necessary  that  for  every  decimal  place  in  the  root, 
there  should  be  three  decimal  places  in  the  power.  Therefore 
we  must  begin  at  the  place  of  units,  and  separate  the  number 
both  to  the  right  and  left  into  periods  of  three  figures  each.  If 
these  do  not  come  out  even  in  the  decimals,  they  must  be  sup- 
plied by  annexing  zeros  to  the  right. 

9.  What  is  the  approximate  third  root  of  25732.75  ? 

10.  What  is  the  approximate  third  root  of  23.1762  ? 

11.  What  is  the  approximate  third  root  of  12f  .'^ 

12.  What  is  the  approximate  third  root  of  m  ?  ' 

13.  What  is  the  approximate  third  root  of  ||  ? 

14.  What  is  the  approximate  third  root  of  /^  ? 

XXXIII.      Questions  producing  Pure  Equations  of  the   Third 

Degree, 

1.  A  man  wishes  to  make  a  cellar,  that  shall  contain  31104 
cubic  feet ;  and  in  such  a  form,  that  the  breadth  shall  be  twice 
the  depth,  and  the  length  I J  the  breadth.  What  must  be  the 
length,  breadth,  and  depth  .'* 

Let  the  depth  =  x, 

the  breadth      =  2  a?, 

and  the  length  =  —. 

The  whole  content  will  be  ^ 

^  ccX2a?X  ?^=:  31104 
3 


I^'=  31104 


14 


162  Ahehra,  XXXIII. 


Algebra, 

16  a?' =  93312 

a;'  =    5832 

a?  =    18  = 

depth 

2a?  =    36  = 

breadth 

S-f  =    48  = 
3 

length. 

2.  There  are  two  men  whose  ages  are  to  each  other  as  5  to 
4,  and  the  sum  of  the  third  powers  of  their  ages  is  137781. 
What  are  their  ages  ? 

Let  X  =.  the  age  of  the  elder 

then =  the  age  of  the  younger. 

o 

125 
0?'  =  91,125 

0?  =  45 

1^  =  36. 
5 

Ans.  Elder  45  years,  and  younger  36. 

3.  A  man  wishes  to  make  a  cubical  cistern  that  shall  con- 
tain 100  gallons.  What  must  be  the  length  of  one  of  its 
sides  ? 

4.  A  bushel  is  2150f  cubic  inches.  What  must  be  the  size 
of  a  cubical  box  to  hold  1  bushel  ? 

5.  What  must  be  the  size  of  a  cubical  box  to  hold  2 
bushels  ? 

6.  What  must  be  the  size  of  a  cubical  box  to  hold  8  bushels  ? 

7.  Find  two  numbers,  such  that  the  second  power  of  the 
greater  multiplied  by  the  less  may  be  equal  to  448 ;  and  the 
second  power  of  the  less  multiplied  by  the  greater,  may  be 
392.? 


XXXIV.  Affected  Equations,  1 63 

8.  A  man  wishes  to  make  a  cistern  which  shall  hold  500 
gallons,  in  such  a  form  that  the  length  shall  be  to  the  breadth 
as  5  to  4,  and  the  depth  to  the  length  as  2  to  5.  Required 
the  length,  breadth,  and  depth. 

Note.     The  wine  gallon  is  231  cubic  inches. 

9.  A  man  wishes  to  make  a  box  which  shall  hold  40  bushels, 
in  such  form  that  the  length  shall  be  to  the  breadth  as  4  to  3, 
and  the  depth  to  the  breadth  as  2  to  3.  Required  the  length, 
breadth,  and  depth  ? 

10.  A  man  bought  a  piece  of  land  for  house  lots,  the  breadth 
of  which  was  to  its  length  as  3  to  28 ;  and  he  gave  as  many 
dollars  per  square  rod,  as  there  were  rods  in  the  length  of  the 
piece.  The  whole  price  was  $63,504.  Required  the  length 
and  breadth. 

11.  A  man  agreed  to  sell  a  stack  of  hay  for  10  times  as  many 
dollars  as  there  were  feet  in  the  length  of  one  of  the  longer 
sides.  On  measuring  it,  the  length  was  to  the  breadth  as  6  to 
5,  and  the  breadth  and  height  were  equal.  Moreover  it  was 
found  that  it  came  to  as  many  cents  per  cubic  foot  as  there 
were  feet  in  the  breadth.  Required  the  dimensions  of  the 
stack. 

XXXIV.     Affected  Equations  of  the  Second  Degree, 

When  an  equation  of  the  second  degree  consists  only  of 
terms  which  contain  the  second  power  of  the  unknown  quanti- 
ty, and  of  terms  entirely  known,  they  may  be  solved  as  above. 
But  an  equation  of  the  second  power,  in  order  to  be  complete, 
must  contain  both  the  first  and  second  powers  of  the  unknown 
quantity,  and  also  one  term  consisting  entirely  of  known  quan- 
tities.    These  are  sometimes  called  affected  equations. 

1.  There  is  a  field  in  the  form  of  a  rectangular  parallelo- 
gram, whose  length  exceeds  its  breadth  by  16  yards',  and  it 
contains  960  square  yards.     Required  the  length  and  breadth. 

Let  X  z=z  the  breadth  ; 

then  X  -\-  16  =z  the  length  ; 

and  x^  -\-  16  a?  r=  the  number  of  square  yards. 

Hence  a:' 4- 16  a;  =^960. 


164  Algebra,  XXXIV. 

In  order  to  solve  this  equation,  it  is  necessary  to  make  the 
first  member  a  perfect  second  power. 

Observe  that  the  second  power  of  the  binomial  a;  +  «,  is  a?*, 
-\-2ax  -^a^,  which  consists  of  three  terms. 

Now  if  we  compare  this  with  the  first  member  *^  -f  16  a?,  we 
find 

2ax  =  16  X  ' 

which  gives        2  a  =  16 
and  a  =  8 

a' =  64 
{x-i-  8)  {x  +  S)=.x^-^l6x  +  64. 

Hence,  if  to  a?*  +  16  a:  we  add  64,  which  is  the  second  power 
of  one  half  of  16,  the  first  member  will  be  a  perfect  second 
power,  but  it  will  be  necessary  to  add  the  same  quantity  to  the 
second  member,  in  order  to  preserve  the  equality.  The  equa- 
tion then  becomes  » 

0^2  -|_  16  a?  +  64  =  960  -f  64  =  1024. 

Taking  the  root  of  both  members 

a?  +  8  ==  dz  (1024)*z=  32. 
By  transposition     x=z  —  8  rt  32. 

It  has  been  already  remarked  that  the  2nd  root  of  every  posi- 
tive quantity,  may  be  either  positive  or  negative,  because  — 

a  X  —  ft  =  +  «'  as  well  as  +  a  x  +  a  =  +  «^     The  double 
sign  =b  is  read  plus  or  minus. 

In  the  preceding  examples,  the  conditions  of  the  question 
have  always  determined  which  was  to  be  used.  But,  in  the 
present  instance,  the  work  not  being  completed  when  the  root 
is  taken,  we  must  give  it  both  signs,  and  when  the  values  of  a: 
are  found  for  both  signs,  the  conditions  will  finally  show  which 
is  to  be  used. 

a;  4-  8  =  -t  32. 


XXXIV.  Affected  Equations,  165 

If  we  use  the  sign  -(-•?  we  have 

cc  =  24 

and  a;  -f  16  =  40. 

This  gives  the  length  40  yards  and  the  breadth  24.  These 
numbers  answer  the  conditions  of  the  question. 

If  v*^e  use  the  sign  — ,  we  have 

a:  =  —  40 

a; -I-  16  =  —-24. 

These  numbers  will  not  satisfy  the  conditions  of  the  question, 
but  they  will  answer  the  conditions  of  the  equation,  as  will  be 
seen  by  putting  them  into  the  first  equation. 

—  40  X  —40  +  16  X  —40  =  960. 

2.  A  certain  company  at  a  tavern  had  a  reckoning  of  143 
shillings  to  pay  ;  but  4  of  the  company  being  so  ungenerous  as 
to  slip  away  without  paying,  the  rest  were  obliged  to  pay  1 
shilling  apiece  more  than  they  would  have  done,  if  all  had  paid. 
What  was  the  whole  number  of  persons  ? 

Let  X  =  the  number  of  persons  at  first ;  : 

then  X  —  4  =  the  number  after  4  have  departed  ; 

143 


X 


=  the  number  of  shillings  each  should  have  paid  ; 


and  —   the  number  of  shillings  actually  paid  by 

each. 

By  the  conditions 

143    ,    J  _     143 

X  X  — 4 

Clearing  of  fractions 

143  a;  4- a;'' —  572  —  4  a:  =  1 43  a: 
By  transposition 

a?'  —  4  a?  =  572. 


166  Algebra,  XXXIV. 

This  equation  is  similar  to  the  last,  except  in  this,  the  se- 
cond term  of  the  first  member  has  the  sign  — . 

Here  we  must  observe  that  the  second  power  of  the  binomial 
X  —  «,  is  a?'  —  2a  X  -\-  a^,  the  same  as  that  of  a?  +  a  with  the 
exception  of  the  sign  of  the  second  term. 

In  this  equation,  as  before,  we  find  two  terms  of  the  second 
power  of  a  binomial  ;  if  we  can  find  the  other  term  we  can 
easily  solve  the  question.  ^'^  '^-^ 

It  may  be  found  as  follows, 

x'  —  x^ 

2ax  =z  —  4x 

2  a  =  —  4 

i^hich  gives  a  =  —  2 

and  a'  =  4 

Adding  4  to  both  members  of  the  equation  it  becomes 

x'^4x-\'4  =  572  +  4=:  576. 

Since  —  2  in  this  corresponds  to  a,  the  root  of  the  first  mem- 
ber is  a?  —  2.  In  fact,  {x  —  2y  =:  x  —  4  a:  +  4.  The  root  of 
576  is  24. 

Hence 

.X  —  2  =  rfc  24 

a?  =  2  rb  24. 

The  two  values  of  a:  are  26  and  —  22.  The  former  only  an- 
swers the  conditions  of  the  question. 

Proof,  If  the  whole  number,  26,  had  paid  their  shares,  each 
would  have  paid  VV  ==  ^i  shillings.  But  22  only  paid,  conr 
sequently  each  paid  VV  ^^  ^i  shillings. 

3.  There  are  two  numbers,  whose  difference  is  9,  and  whose 
sum  multiplied  by  the  greater  produces  266.  What  are  those 
numbers  f 

Let  X  =  the  greater ; 

then  X  —  9  =r  the 


2x  —  9  =  their  sum.  ^ 


XXXIV.  Affected  Equations.  167 

By  the  conditions 

cr(2a?  — 9)=266 
2  a?*  —  9  a?  =  266 


--'r  = 

133. 

'we  use 

the 

general  fonnula  as  before, 

we 

have 

a^  = 

:x' 

•*  ^ 

2aa;  = 

2a  = 

a  = 

a'  = 

__9x 

2 

_9 
2 

9 
4 

.  81 

16 

Completing  the  second  power,  the  equation  becomes 

2209 


2 

^  16        '*"'  "^  16           16 

Taking  the  root  of  both  members 

* 

4              4 

4          4 

which  gives 

.=  55=14 
4 

and 

»=-?=-.*, 

a?— -9  =  5 

also 

J?— 9=—18f 

m  Algebra.  XXXIV. 

Both  values  will  answer  the  conditions  of  the  question  ;  for 

14  +  5  =  19 
and  19  x  14  =266 

also  —  9J  ^-  (_  18J)  =  —  28 

and  --_  28  X  —  9J  =  266. 

In  all  the  above  examples,  after  the  question  was  put  into  equa- 
tion, the  first  thing  done,  was  to  reduce  all  the  terms  contain- 
ing a?^  to  one  term,  and  those  containing  x  into  another,  and 
to  place  them  in  one  member  of  the  equation,  and  to  collect  all 
the  terms  consisting  entirely  of  known  quantities  into  the  other. 
This  must  always  be  done.  Moreover  x^  must  have  the  sign 
+  and  its  coefficient  must  be  1.  The  equation  will  then  be  in 
the  following  form. 

x^  -{-  p  X  =:  q. 

p  and  q  being  any  known  quantities  and  either  positive  or 
negative. 

Every  equation,  however  complicated,  consisting  of  terms 
which  contain  x^,  and  x,  and  known  quantities  may  be  reduced 
to  this  form. 

Let  the  equation  be  +  ^  '■ 

5  Ax  —  2 

Clearing  of  fractions  it  becomes 

140  a?  —  12  x*  —  70  +  6  >r  =  75  —  5  x\ 
Transposing  and  uniting  terms 

146  a?— 7  a;' =  145 
Changing  all  the  signs  in  both  members 
7a:2__  146a?  =  —  145 
Dividing  by  7  (the  coefficient  of  a:') 

a  _  146  a?  __  145 

"^7  "7" 


XXXIV.  Affected  Equations.  169 

TT  146       ,  145 

Here  p  =.  — — -  and  o  =  — 

T      .  7 

To  solve  the  equation 

We  consider  cc*  and  ^  a?  as  two  terms  of  the  second  power 
of  the  binomial  x  -\-  ain  which 

2ax  =zpx     ' 
2a  =^p 

a  =  P 

2 

4 

Hence  the  binomial  x  -\-  a  is  equal  to  x  -{-?-,  and  the  third 

2 

term  of  the  second  power  is  ii.     In  fact 

Therefore  the  first  member  of  the  above  equation  may  be 
rendered  a  complete  second  power,  of  which  cc  -j-  ^  is  the 

root,  by  adding  to  it  ^.     The  same  quantity  must  be  added  to 

4 

the  second  member,  to  preserve  the  equality. 
The  equation  then  becomes 


4                4 

raking  the  root  of  both  members 

'H-H'+^Y 

'=-!-('+?)* 

16 

if^  Algdra.  XXXIV. 

From  the  above  observations  we  derive  the  following  general 
rule  for  the  solution  of  equations  which  contain  the  first  and 
second  powers  of  the  unknown  quantity. 

1st.  Prepare  the  equation,  by  collecting  all  the  terms  contain- 
ing the  first  and  second  powers  of  the  unknown  quantity  into  tJte 
first  member,  and  all  tlie  terms  consisting  entirely  of  known  quan- 
tities into  the  other  member.  Unite  all  the  terms  containing  the 
second  power  into  one  term,  and  all  containing  the  first  power 
into  another.  If  the  sign  before  the  term  containing  the  second 
power  of  the  unknown  quantity  be  not  positive,  make  it  so  by 
changing  all  the  signs  of  both  members.  If  the  coefficient  of 
this  term  is  not  1,  make  it  so  by  dividing  all  the  terms  by  its  coef- 
ficient. 

2d.  Make  the  first  member  a  complete  second  power.  This  is 
done  by  adding  to  both  members  the  second  power  of  half  the  coeffi- 
cient of  X  (or  of  the  first  power  of  the  unknown  quantity.) 

3d.      Take  the  root  of  both  members. 

The  root  of  the  first  member  will  be  a  binomial,  the  first  term  of 
which  will  be  the  unknown  quantity,  and  the  second  will  be  half  the 
coefficient  of  x  as  found  above.  The  root  of  the  second  member 
must  have  the  double  sign  =t. 

4th.  Transpose  the  term  consisting  of  known  quantities  from  the 
first  to  the  second  member,  and  the  value  of  x  will  be  found. 

4.  A  and  B  sold  130  ells  of  silk  (of  which  40  ells  were  A's 
and  90  B's)  for  42  crowns.  Now  A  sold  for  a  crown  one  third 
of  an  ell  more  than  B  did.  How  many  ells  did  each  sell  for  a 
crown  ? 

Let  X  =  the  number  of  ells  B  sold  for  a  crown ;  then  x  + 
J  =  the  number  A  sold  for  a  crown ; 

?2  —  the  price  of  90  ellsj 

X 

=  the  price  of  40  ells. 

^+i2_  =  42 

X         ^+ J 


XXXIV.  Affected  Eguaiions,  171 

90  -f  1*^  =:  42  a; 
^i^  +  i 

90  a: -h  30 +  40  a?  =  42  07* -I-  14  a? 
116a'  — 42a?*  =  —  30 
Changing  signs  42  a?''  —  1 16  a?  =  30 

Dividing  by  42  x'  —  il^f  =  ^ 

^    ^  42  42 

I        Reducmff  fractions   a?*  — f  =  — 

I  ^  21  7 

1       To  complete  the  second  power  of  the  first  member,  take  one 
^   half  of  — If,  which  is  —  |f,  and  add  its  second  power  to 
both  members. 


^__58^       J41^_5       _841  _^15     ,    _B4£  __  1156 
21      '^  2iy       7^  2ir~  2ir  "^W""    21/ 
Taking  the  root  of  both  members, 

29         ,    34 

a? — —  =  db  — 
21  21 

29    ,     34 
21         21 


Which  give         a?  =  — 
^  21 


and 


5 

^"~       21 
The  first  value  only  will  answer  the  conditions. 

Ans.  B  sold  3  ells  for  a  crown,  and  A  3J. 

The  learner  may  observe,  that  in  raising  |f  to  the  second 
power,  I  multiplied  the  numerator  into  itself,  but  expressed  the 
power  of  the  denominator  by  an  exponent.  This  saved  some 
work  in  this  example.  It  may  always  be  done  when  the  num- 
ber in  the  right  hand  member  can  be  reduced  to  a  fraction 
with  the  same  denominator  as  the  number  added.  In  this  case 
4  could  be  reduced  to  21ths.     The  f  was  reduced  thus 


172  Algebra.  XXXIV. 

5x3  _   15  X  21   __  315 
7X3         21  X  21         21)2 

When  the  second  member  is  a  whole  number,  it  can  be  re- 
duced to  a  fraction  with  any  denominator  ;  consequently  this 
form  may  be  used. 

5.  A  man  bought  a  certain  number  of  sheep  for  80  dollars  ; 
if  he  had  bought  4  more  for  the  same  money,  they  would  have 
come  to  him  1  dollar  apiece  cheaper.  What  was  the  number 
of  sheep  ^ 

6.  A  merchant  sold  a  quantity  of  brandy  for  £39  and  gained 
as  much  per  cent,  as  the  brandy  cost  him.  How  much  did  it 
cost  him  .'' 

^  Let  X  =  the  cost. 

then =  the  rate  per  cent. 

100  ^ 

and  — -  =  the  eain. 
100  ^ 

also  39  —  X  z=z  the  gain. 

7.  Two  persons,  A  and  B,  talking  of  their  money,  says  A  to 
B,  if  I  had  as  many  dollars  as  I  have  shillings,  I  should  have  as 
much  money  as  you  ;  but  if  I  had  as  many  shillings  as  their 
number  multiplied  by  itself,  I  should  have  three  times  as  much 
money  as  you,  and  63  shillings  over.  How  much  money  had 
each  ^ 

8.  A  colonel  has  a  battalion  of  1200  men,  which  he  would 
draw  up  in  a  solid  body  of  an  oblong  form,  so  that  each  rank 
may  exceed  each  file  by  59  men.  What  numbers  must  he 
place  in  rank  and  file  ^ 

9.  A  grazier  bought  as  many  sheep  as  cost  him  £60 ;  out  of 
which  he  reserved  15,  and  sold  the  remainder  for  £54,  gaining 
2  shillings  a  head  by  them.  How  many  sheep  did  he  buy,  and 
what  was  the  price  of  each  ? 

10.  A  person  bought  two  pieces  of  cloth  of  different  sorts  ; 
of  which  the  finer  cost  4  s.  a  yard  more  than  the  other.  For 
the  finer  he  paid  £18  ;  but  for  the  coarser,  which  exceeded 
the  finer  in  length  by  2  yards,  he  paid  only  £16.     How  many 


XXXIV,  Jlffeded  Equations,  173 

yards  were  there  in  each  piece,  and  what  was  the  price  of 
each  f 

11.  A  labourer  dug  two  trenches,  one  of  which  was  16  yards 
longer  than  the  other,  for  $77.60  ;  and  the  digging  of  each 
cost  as  many  dimes  per  yard,  as  there  were  yards  in  length. 
What  was  the  length  of  each  ? 

12.  There  are  two  square  buildings,  that  are  paved  with 
stones  each  a  foot  square.  The  side  of  one  building  exceeds 
that  of  the  other  by  12  feet,  and  both  their  pavements  taken 
together  contain  2120  stones.  What  are  the  lengths  of  them 
separately. 

13.  A  man  bought  two  sorts  of  linen  for  f  13^.  A  yard  of 
the  finer  cost  as  many  shillings  as  there  were  yards  of  the  finer. 
Also  30  yards  of  the  coarser,  (which  was  the  whole  quantity,) 
were  at  such  a  price,  that  7  yards  cost  as  much  as  a  yard  of  the 
finer.  How  many  yards  were  there  of  the  finer,  and  what  was 
the  value  of  each  piece  ? 

14.  Two  partners  A  and  B  gained  £18  by  trade.  A's  mo- 
ney was  in  trade  12  months,  and  he  received  for  his  principal 
£uid  gain  £26.  Also  B's  money,  which  was  £30,  was  in  trade 
16  months.     What  money  did  A  put  into  trade  ?    '  • 

15.  The  plate  of  a  looking  glass  is  18  inches  by  12,  and  is 
to  be  framed  with  a  frame,  all  parts  of  which  are  of  equal  width, 
and  the  area  of  the  frame  is  to  be  equal  to  that  of  the  glass. 
Required  the  width  of  the  frame. 

16.  A  and  B  set  out  from  two  towns,  which  were  distant  247 
miles,  and  travelled  the  direct  road  till  they  met.  A  went  9 
miles  a  day  ;  and  the  number  of  days,  at  the  end  of  which  they 
met,  was  greater  by  3  than  the  number  of  miles  which  B  went 
in  a  day.     How  many  miles  did  each  go  ? 

17.  A  set  out  fi'om  C  towards  D,  and  travelled  7  miles  per 
day.  After  he  had  gone  32  miles,  B  set  out  from  D  towards  C, 
and  went  every  day  yV  of  the  whole  journey  ;  and  after  he  had 
travelled  as  many  days  as  he  went  miles  in  one  day,  he  met  A. 
What  is  the  distance  between  the  places  C  and  D  ? 

In  this  case  both  values  will  answer  the  conditions  of  the 
question. 

15* 


174  Algebra,  XXXIV. 

18.  A  man  had  a  field,  the  length  of  which  exceeded  the 
breadth  by  5  rods.  He  gave  3  dollars  a  rod  to  have  it  fenced, 
which  amounted  to  1  dollar  for  every  square  rod  in  the  field. 
What  was  the  length  and  breadth,  and  what  did  he  give  for 
fencing  it  f 

19.  From  two  places  at  a  distance  of  320  miles,  two  persons, 
A  and  B,  set  out  at  the  same  time  to  meet  each  other.  A  tra  • 
veiled  8  miles  a  day  more  than  B,  and  the  number  of  days  in 
which  they  met  was  equal  to  half  the  number  of  miles  B  went  in 
a  day.    How  many  miles  did  each  travel,  and  how  far  per  day  ^ 

20.  A  man  has  a  field  15  rods  long  and  12  rods  wide,  which 
he  wishes  to  enlarge  so  that  it  may  contain  just  twice  as  much ; 
and  that  the  length  and  breadth  may  be  in  the  same  propor- 
tion.    How  much  must  each  be  increased  ? 

In  this  example,  the  root  can  be  obtained  only  by  approxi- 
mation. 

21.  A  square  court  yard  has  a  rectangular  gravel  walk 
round  it.  The  side  of  the  court  wants  2  yards  of  being  6 
times  the  breadth  of  the  gravel  walk ;  and  the  number  of 
square  yards  in  the  walk  exceeds  the  number  of  yards  in  the 
periphery  of  the  court  by  164.  Required  the  area  of  the 
court  ^ 

All  equations  of  the  second  degree  may  be  reduced  to  one 
of  the  following  forms. 

1 .  x^  -{-  px  =z  q 

2.  x^  — p  X  z=  q 

3.  x^-\-px=: — q 

4.  a?"  — p  a?  =  —  q. 

After  the  equation  has  been  brought  to  one  of  these  forms, 
it  may  be  solved  by  one  of  the  following  formulas,  which  are 
numbered  to  correspond  to  the  equations  from  which  they  are 
derived. 

1.  x  =  -|±(5  +  f)* 

2.  x=+|±(?  +  f)4 


XXXV.  Affected  Equations.  175 

4  4 

The  first  equation  and  the  first  formula  are  suflicient  for  the 
whole,  if  p  and  q  are  supposed  to  be  positive  or  negative  quan- 
tities. 

21.  There  are  two  numbers  whose  difference  is  11|,  and 
whose  product  is  equal  to  4  times  the  larger  minus  9.  What 
are  the  numbers  ^ 

Let  X  —  the  larger ;  • 

then  X  —  1 1  f  =  the  smaller. 

a?'  —  llf«  =  4a;  — 9 

This  equation  is  in  the  form  of  x^  —  px  =  —  q/ia  which 

78     p         78     p"       6084       i  ^ 

»  =  — ,  ^  =  — J  ^  =   and  a  =  9. 

^         5      2         10     4         100  ^ 

J?  =  II  ±  ( VoV  —9)*  =  f I  ±  ( VoV)*  =  '7.8  i  7.2, 
Or  we  may  use  the  first  formula,  then  H 

==  —  1?  P  —  ^1^  tz=i  ^^  and     =  —  9 
^  ~        5  '  2  lO'  4         100 '    "    ^  ~" 


X  =  it  ±  (  VoV  —  9)*  =  U  =fc  (  VVV)*  =  7.8  i  7.2. 
Both  values  of  a?,  being  positive,  will  answer  the  conditions 
of  the  question. 

Jim.  By  the  first  value  the  larger  number  is  15  and  the 
smaller  3|.  By  the  second  value  of  x,  the  larger  is  |,  and  the 
smaller  —  11. 

Let  the  learner  s.olve  some  of  the  preceding  questions  by  the, 
formula. 

XXXV.  We  shall  now  demonstrate  that  every  equation  of 
the  second  degree,  necessarily  admits  of  two  values  for  the  un- 
known quantity,  and  only  two. 


176  Algehra.  XXXV. 

Let  us  take  the  general  equation. 

x^  -{-  p  oc  =^  q. 

This,  we  have  seen,  may  represent  any  equation  whatever  of 
the  second  degree,  p  and  q  being  any  known  quantities  and 
either  positive  or  negative.     If  p  =  0  the  equation  beconies 

x''  =  q, 

which  is  a  pure  equation  or  an  equation  with  two  terms. 

If  we  make  the  first  member  of  the  equation  x^  -]-px  =.q^ 
a  complete  second  power,  by  the  above  rules,  it  becomes 

x^^px^jt=.q-\-£ 
4  4 

or       .       (x  +  |/=j  +  |!. 


Make 

m' 

■-q^ 

P 

4 

then 

m 

= 

(9  + 

P\i 
4^ 

Then  we 

have 

{X 

+I-: 

2 

transposing  m^       {x  +  ^)  —  m^  =  0. 

The  first  member  of  this  equation  is  the  difference  of  two 
second  powers,  which.  Art.  XIII,  is  the  same  as  the  product  of 
the  sum  and  difference  of  the  numbers. 

The  sum  is  a?  -f-  ^  4-  w,  and  the  difference  \^  x  -\-  JL  —  m, 

2   ^  2 

and  their  product  is 

In  this  equation,  the  first  member  consists  of  two  factors, 
and  the  second  is  zero.  Now  the  first  member  of  the  above 
equation  will  be  equal  to  zero,  if  either  of  its  factors  is  equal 


1 


XXXV.  Affected  Equations.  177 

to  zero.     For  if  any  number  be  multiplied  by  zero,  the  product 
is  zero. 

Making  the  first  factor  equal  to  zero, 


x-^£-.-^m=:0 

gives 

.  =  -Z  +  „,. 

Making 

x+^+m  =  0 

gives 

X  =  —  £-  —  m. 

2 

Either  of  these  values  of  a?  must  answer  the  conditions  of  the 
equation. 

N.  B.  Though  either  value  answers  the  conditions  separate- 
ly, they  cannot  be  introduced  together,  for  being  different, 
their  product  cannot  be  x^. 

Instead  of  m  put  its  value,  and  the  values  of  x  become 

which  are  the  values  we  had  obtained  above.     (This  demon- 
stration is  essentially  that  of  M.  Bourdon.) 

Discussion. 
Let  us  take  again  the  general  equation. 

Since  the  expression  contains  a  radical  quantity,  that  is,  a 
quantity  of  which  the  root  is  to  be  found,  in  order  to  be  able 
to  find  the  value  of  it,  we  must  be  able  to  find  the  root  either 
exactly  or  by  approximation.     Now  there  is  one  case  in  which 


178  Algebra.  XXXV. 

it  is  impossible  to  jSnd  the  root.     It  is  when  q  is  negative  and 

2  2 

greater  than    P-,    In  which  case  the  expression  o  -{-  ^  is  ne- 
4  4 

gative  ;  and  it  has  been  shown  above,  that  it  is  impossible  to 
find  the  root  of  a  negative  quantity.  In  all  other  cases  the 
value  of  the  equation  may  be  found. 

In  all  cases  if  q  is  positive,  the  first  value  will  be  positive, 
and  answer  directly  to  the  conditions  of  the  question  proposed. 

For  the  radical  Iq  -f-  ^^  I*  is  necessarily  greater  than  ^,  be-     ^ 
cause  the  root  of  S-  alone  is  £~  ;  therefore   the   expression 

—  ^  ^  i^'"'"'^)        necessarily  of  the  same  sign  as  the 

radical. 
The  second  value  is  for  the  same  reason  essentially  negative, 

for  both  -^  and  lq-\-  ^y  are  negative.    This  value,  though 

it  fulfils  the  conditions  of  the  equation,  does  not  answer  the 
conditions  of  the  question,  from  which  the  equation  was  derived ; 
but  it  belongs  to  an  analogous  question,  in  which  the  x  must  be 
put  in  with  the  sign  —  instead  of  -}-  ;  thus  x^  — px=^  q,  which 

gives  X  =z  E.  i  (s'+  — Ij^  value,  which  differs  from  the 
first  only  by  the  sign  before  -£-. 

If  q  is  actually  negative,  the  equation  becomes 
x^^px-=.  —  q, 
and  the  values  are 

In  order  that  it  may  be  possible  to  find  the  root,  q  must  be 
less  than  ^.     When  this  is  the  case,  the  two  values  are  real. 


XXXV.  Affected  Equations.  179 

Since    1^  — 5' F  is  smaller  than  ^,  it  follows  that   both 

values  are  negative  if  p  is  positive  in  the  equation  ;  that  is,  if 
vT*  +jp  J?  =  —  q,  which  gives 

and  both  positive  if  p  is  negative  in  the  equation,  that  is,  x^  — 
p  jc  =  —  q^  which  gives 

i 


-f^(|-.y 


When  both  values  are  negative,  neither  of  them  answers  di- 
rectly to  the  conditions  of  the  question  ;  but  if  —  a?  be  put  into 
the  original  equation  instead  of  a?,  the  new  equation  will  show 
what  alteration  is  to  be  made  in  the  enunciation  of  the  ques- 
tion ;  and  the  same  values  will  be  found  for  x  as  before,  with 
the  exception  of  the  signs. 

If  in   this  equation  q  is  greater    than  i-,    the   quantity 

§£- —  5'!*  becomes  negative,  and  the  extraction  of  the  root 

cannot  be  performed.     The  values  are  then  said  to  be  imagi- 
nary. 

1.  It  is  required  to  find  two  numbers  whose  sum  is  p,  and 
whose  product  is  q. 

Let  X  =  one  of  the  numbers, 
then  p  —  X  =  the  other. 

x(p  —  ^)  =^  q 
p  X  —  x^  :=:iq; 
Changing  signs         x^  —  p  x  =.  —  q. 

This  example  presents  the  case  above  mentioned,  in  which 
p  and  q  are  both  negative. 


180  Algebra.  XXXV. 

The  value  is 

Suppose^  =15  and  q  =  54. 

z  =  i5  i  /^  _^4U^15^/225-Bl6U 

__15        3^ 

2         2* 

The  values  are  9  and  6,  both  positive,  and  both  answer  the 
conditions  of  the  question.  And  these  are  the  two  num- 
bers required,  for  9  -f-  6  =  15,  9  X  6  =  54.  This  ought  to  be 
so,  for  X  in  the  equation  represents  either  of  the  numbers  in- 
differently. Indeed  whichsoever  x  be  put  for,  p  —  x  will  re- 
present the  other  ;  andp  a:  —  x^  will  be  their  product. 

Agam  letp  =  16  and  g^  =  72. 

Here  ( —  8)^  is  an  imaginary  quantity,  therefore  both  values 
are  imaginary. 

In  order  to  discover  why  we  obtain  this  imaginary  result,  let  us 
first  find  into  what  two  parts  a  number  must  be  divided,  that  the 
product  of  the  two  parts  may  be  the  greatest  possible  quantity. 

In  the  above  example,  p  represents  the  sum  of  the  two  num- 
bers or  parts,  let  d  represent  their  difference,  then 

1-+  —  —  the  greater,  andZ  —  .1  =  the  less.  Art.  IX. 

2^2  ^  2-2 

Their  product  is 

iV_  +  IW^  -- 1\  =£!-!'  Art.  XIII. 

\2         2/   \2         2/         4  4 


8?. 

—  IS 

4 
d  is  greater  than  zero ;  but  when  e?  =  0,  the  expression  becomes 


The  expression  ^  —  —  is  evidently  less  than  2-  so  long  as 


XXXV.  Affected  Equation.  181 

£.  which  is  the  second  powei*  of  ^.  Therefore  the  greatest 
possible  product  is  when  the  two  parts  are  equal. 

2 

In  the  above  example  ^  =  8,  and  E-    =z  64.     This  is  the 
^2  4 

greatest  possible  product  that  can  be  formed  of  two  numbers 
whose  sum  is  16.  It  was  therefore  absurd  to  require  the  pro- 
duct to  be  72  ;  and  the  imaginary  values  of  x  arise  from  that 
absurdity. 

2.  It  is  required  to  find  a  number  such,  that  if  to  its  second 
power,  9  times  itself  be  added,  the  sum  will  be  equal  to  three 
times  the  number  less  5. 

a;'-f  9a?  =  3a?  — 5. 

x^  -\-  6  X  =1  —  5. 

This  equation  is  in  the  form  of  a?*  +  ^  j?  =  —  q,  which 
gives 

Putting  in  the  values  of  p  and  q 

a:=  — 3±(9— 6)*  =  — 3±2. 

The  values  are  —  1  and  —  5,  both  negative.  Consequently 
neither  value  will  answer  the  conditions  of  the  question.  This 
shows  also  that  those  conditions  cannot  be  answered. 

But  if  we  change  the  sign  of  x  in  the  equation,  that  is,  put 
in  —  X  instead  of  a?,  it  becomes 

x^  —  9x=z  —  Sx^5, 
Changing  all  the  signs 

9a:  —  a?'  =  3a?  +  5. 
This  shows  that  the  question  should  be  expressed  thus : 

It  is  required  to  find  a  number,  such,  that  if  from  9  times 
itself,  its  second  power  be  subtracted,  the  remainder  will  be 
equal  to  3  times  the  number  plus  5. 
16 


182  Algebra.  XXXVI. 

The  values  will  both  be  positive  in  this,  and  both  answer  the 
conditions. 

a?^  —  9a?  =  —  3^ — 5 

x^  —  6  0?  =  —  5 

a:  =  3  dz  (9  —  5)*  =  3  ±  2, 

The  values  are  5  and  1  as  before,  but  now  both  are  positive, 
and  both  answer  the  conditions  of  the  question. 

3.  There  are  two  numbers  whose  sum  is  a,  and  the  sum  of 
whose  second  powers  is  6.     It  is  required  to  find  the  numbers. 

Examine  the  various  cases  which  arise  from  giving  different 
values  to  a  and  h.  Also  how  the  negative  value  is  to  be  inter- 
preted.    Do  the  same  with  the  following  examples. 

4.  There  are  two  numbers  whose  difference  is  a,  and  the 
sum  of  whose  second  powers  is  h.     Required  the  numbers. 

5.  There  are  two  numbers  whose  difference  is  a,  and  the 
difference  of  whose  third  powers  is  h.    Required  the  numbers. 

6.  A  man  bought  a  number  of  sheep  for  a  number  a  of  dol- 
lars ;  and  on  counting  them  he  found  that  if  there  had  been  a 
number  b  more  of  them,  the  price  of  each  would  have  been  less 
by  a  sum  c.     How  many  did  he  buy  ^ 

7.  A  grazier  bought  as  many  sheep  as  cost  him  a  sum  «,  out 
of  which  he  reserved  a  number  6,  and  sold  the  remainder  for  a 
sum  c,  gaining  a  sum  d  per  head  by  them.  How  many  sheep 
did  he  buy,  and  what  was  the  price   of  each  ^ 

8.  A  merchant  sold  a  quantity  of  brandy  for  a  sum  a,  and 
gained  as  much  per  cent,  as  the  brandy  cost  him.  What  was 
the  price  of  the  brandy  ^ 


XXXVI.     Of  Powers  and  Roots  in  General. 

Some  explanation  of  powers  both  of  numeral  and  literal 
quantities  was  given  Art.  X.  The  method  of  finding  the  roots 
of  the  second  and  third  powers,  that  is,  of  finding  the  second 
and  third  roots  of  numeral  quantities,  has  also  been  explained; 
and  their  application  to  the  solution  of  equations.     But  it  is 


XXXVI.  Of  Powers  and  Roots  in  General.  183 

frequently  necessary  to  find  the  roots  of  other  powers,  as  well 
as  of  the  second  and  third,  and  of  literal,  as  well  as  of  numeral 
quantities.  Preparatory  to  this,  it  is  necessary  to  attend  a 
little  more  particularly  to  the  formation  of  powers. 

The  second  power  of  a  is  a  X  «  =  «^' 

The  fifth  power  of  a  is  «  x  «  X  «  X  «  X  «  =  a^ 

If  a  quantity  as  a  is  multiplied  into  itself  until  it  enters  m 
times  as  a  factor,  it  is  said  to  be  raised  to  the  /nth  power,  and  is 
expressed  a"*.  This  is  done  by  m  —  1  multiplications  ;  for  one 
multiplication  as  a  X  «  produces  a^  the  second  power,  two 
multiplications  produce  the  third  power,  &c. 

We  have  seen  above  Art.  X.  that  when  the  quantities  to  be 
multiplied  are  alike,  the  multiplication  is  performed  by  adding 
the  exponents.  By  this  principle  it  is  easy  to  find  any  power 
of  a  quantity  which  is  already  a  power.     Thus 

The  second  power  of  a'  is  a^  x  a'  =  «^^  =  a^. 

The  third  power  of  a"  is  a'  X  a'  X  a'  ==^  a'H-V  =  a\ 

The  second  power  of  a"*  is  a"^  x  «"*  =  am+m  =as  ^^m.^ 

The  third  power  of  a™  is  «»"  X  a""  X  a'"  =  a»"+™+"^  —  a^™ . 

The   mth   power  of  aMs   a^   X   «^  X   a^  X  «^   X 

—  ^2+2+2+2+ ^  until  a^  is  taken  m  times  as  a  factor,  that 

is,  until  the  exponent  2  has  been  taken  m  times.  Hence  it  is 
expressed  a^*". 

The  nth  power  of  a™  is  a»"  x  «"*  X  a"" =  fjm-\^^m+  .  . . 

until  m  is  taken  n  times,  and  the  power  is  expressed  a"^". 

N.  B.     The  dots in  the  two  last  examples  are  used  to 

express  the  continuation  of  the  multiplication  or  addition,  be- 
cause it  cannot  come  to  an  end  until  m  in  the  first  case,  and  n 
in  the  second,  receive  a  determinate  value. 

In  looking  over  the  above  examples  we  observe  ; 

1st.  That  the  second  power  of  a^  is  the  same  as  the  third 
power  of  a^,  and  so  of  all  others.  ' 


184  Algebra,  XXXVI. 

2.  That  in  finding  a  power  of  a  letter  the  exponent  is  added 
until  it  is  taken  as  many  times  as  there  are  units  in  the  expo- 
nent of  the  required  power.  Hence  any  quantity  may  he  raised 
to  any  power  by  multiplying  its  exponent  by  the  exponent  of  the 
power  to  which  it  is  to  be  raised. 

The  5th  power  of «'  is  a'x'  =  a'K 

The  3d  power  of  a'  is  a'^^  =  «^\  &c. 

The  power  of  a  product  is  the  same  as  the  product  of  that 
power  of  all  its  factors. 

The2dpowerof3a6is3a6  X  3a6=:9  aH\ 

The  3d  power  of  2  a'^^'is  2  «'  6'  X  2  a' J'  X  2  a'  6'  =:  8  a'  b\ 

Hence,  when  a  quantity  consists  of  several  letters,  it  may  be  rais- 
ed to  any  power  by  multiplying  the  exponents  of  each  letter  by  the 
exponent  of  the  power  required  ;  and  if  the  quantity  has  a  numeral 
coefficient,  that  must  be  raised  to  the  power  required. 

The  powers  of  a  fraction  are  found  by  raising  both  numera- 
tor and  denominator  to  the  power  required  ;  for  that  is  equiva- 
lent to  the  continued  multiplication  of  the  fraction  by  itself. 

1  What  is  the  5th  power  of  S  a^  b"^  m  ? 

2  What  is  the  3d  power  of  ^^' .? 

Powers  of  compound  quantities  are  found  like  those  of  sim- 
ple quantities,  by  the  continued  multiplication  of  the  quantity 
into  itself  The  second  power  is  found  by  multiplying  the 
quantity  once  by  itself.  The  third  power  is  found  by  two  mul- 
tiplications, &c. 

The  powers  of  compound  quantities  are  expressed  by  enclos- 
ing the  quantities  in  a  parenthesis,  or  by  drawing  a  vinculum 
over  them,  and  giving  them  the  exponent  of  the  power.  The 
third  power  of  «  -}-  2  6  —  c  is  expressed  (a  -j-  2  6  —  cf  ;  or 


a  -f  2  6  —  c. 

The  powers  are  found  by  multiplication  as  follows 


XXXVL  Of  Powers  and  Roots  in  General.  185 

a  +  2h—c 
aJ^^h  —  c 

a^  -\-2ab  —  ac 

—  ac  —  2&C  +  C* 


a'^4ab-\-4b'  —  2ac--4bc-}-e'  =  {a  +  2b  —  cy 
a-\-2b  —  c 

a^  +  4an+4a¥—2a^c  —  4abc-i-ac' 

2a'b-\-Sab'+8¥  ■^4abc—8b'c-\-2b c" 

-_a»c  —4abc—4b'c+2ac'-\'4bc^'--c' 

a^  ^  Q  a"  b  -^  12  a  6*  +  86'—  3  a'c— 12  a6c— 12  6'c 
+  3ac'  +  66c'  —  c'  =  (a  +  26—  c)'. 

If  the  third  power  be  multiplied  by  a  -{-2  b  —  c,  it  will  pro- 
duce the  fourth  power. 

3.  What  is  the  second  power  of  3  c  +  2  e^  ? 

r 

4.  What  is  the  third  power  of  4  a  —  6  c  ? 

5.  What  is  the  fifth  power  of  a  —  b? 

6.  What  is  the  fourth  power  of  2  a^  c  —  c^  ? 

In  practice  it  is  generally  more  convenient  to  express  the 
powers  of  compound  quantities,  than  actually  to  .find  them  by 
multiplication.  And  operations  may  frequently  be  more  easily 
performed  on  them  when  they  are  only  expressed. 

(a  4-  by  X  («  +  by  =  («  +  b)^'  =  (a  +  by 
(3  a—  5  c)'  X  (3  a  —  5  cV  =  (3  a—  5  c  )'. 
16* 


186  Alg^ra.  XXXVI. 

That  is,  when  one  power  of  a  compound  quantity  is  to  he  multi- 
plied by  any  power  of  the  same  quantity,  it  may  be  expressed  by  add- 
ing the  exponents,  in  the  same  manner  as  simple  quantities. 

The  2d  power  of  {a  +  bf  is  {a  +  bf  X  («  +  bf 

=  (« _|.5)3+3  :=(«-!-  by^'  =  (« +  by. 

The  3d  power  of  (2  a  —  ^)Ms 

(2«  — rfy+4+4  =:(2a'^d)*x'z={2a'—dy\ 

That  is,  any  quantity,  which  is  already  a  power  of  a  compound 
quantity,  may  be  raised  to  any  power  by  multiplying  its  exponent 
by  the  exponent  of  the  power  to  which  it  is  to  be  raised. 

7.  Express  the  2d  power  of  (3  6  —  e)*. 

8.  Express  the  3d  power  of  (a  —  c  +  2  dy. 

9.  Express  the  7th  power  of  (2  a'  — 4  c')'. 

Division  may  also  be  performed  by  subtracting  the  exponents 
as  in  simple  quantities. 

(3  ff  —  by  divided  by  (3  a  —  by  is 

^Sa  —  b)'-'z=:{2a-^by 

10.  Divide  (7  m  +  2  c)'  by  (7  m  -L.  2  c)^ 

If  (a  +  by  is  to  be  multiplied  by  any  quantity  c,  it  may  be 
expressed  thus :  c  (a  -{-  by.  But  in  order  to  perform  the  ope- 
ration, the  2d  power  of  a  -\-b  must  first  be  found. 

c  {a  +by  =z  c  {a^  -{-  2  ab  +  b^)  =  a'  c  -\-2  ab  c  -{-  b'  c 

If  the  operation  were  performed  previously,  a  very  erroneous 
result  would  be  obtained ;  for  c  (a  +  by  is  very  different  from 
(a c  +  6 c)^  The  value  of  the  latter  expression  is  a^  c^  -\- 2  ab 
c'-^b'c\ 

11.  What  is  the  value  of  2  (a  +  Sby  developed  as  above  ? 

12.  What  is  the  value  of  3  5  c  (2  a  —c)'  ? 

13.  What  is  the  value  of  (a  +  3  c')  (3  a  —  2  by  ? 

14.  What  is  the  value  of  (2  a  —  by  (a'  +  b  c)*  ? 


XXX VI.  Of  Powers  and  Roots  in  General.  187 

We  have  had  occasion  in  the  preceding  pages  to  return  from 
the  second  and  third  powers  to  their  roots.  We  have  shown 
how  this  can  be  done  in  numeral  quantities ;  it  remains  to  be 
shown  how  it  may  be  effected  in  literal  quantities.  It  is  fre- 
quently necessary  to  find  the  roots  of  other  powers  as  well  as 
of  the  second  and  third. 

The  power  of  a  literal  quantity,  we  have  just  seen,  is  found 
by  multiplying  its  exponent  by  the  exponent  of  the  power  to 
which  it  is  to  be  raised. 

The  second  power  of  a^s  a^  ^  '  =  a^ ;  consequently  the  se- 
cond  root  of  a^  is  a^  =  ft^ 

The  third  power  of  a^  is  a^  ;  hence  the  third  root  of  a^ 
must  be  «  3  =:  a^. 

m 

The  second  root  of  a*",  then  must  be  a^. 
Proof.    The  second  power  of  a^  jg  ct  2  ^^  a"^^ 

In  general,  the  root  of  a  literal  quantity  may  be  found  by  divid- 
ing its  exponent  by  the  number  expressing  the  root  ;  that  is,  by 
dividing  by  2  for  the  second  root,  by  3  for  the  third  root,  &c. 
This  is  the  reverse  of  the  method  of  finding  powers. 

It  was  shown  above,  that  any  power  of  a  quantity  consisting 
of  several  factors  is  the  same  as  the  product  of  the  powers  of 
the  several  factors.  From  this  it  follows,  that  any  root  of  a 
quantity  consisting  of  several  factors  is  the  same  as  the  pro- 
duct of  the  roots  of  all  the  factors. 

The  third  power  of  a*  b  c^  is  a^  b^  c' ;  the  rfiird  root  of  a^  b^ 
c^  must  therefore  be  a^  b  c^ 

Numeral  coeMcients  are  factors,  and  in  finding  powers 
they  are  raised  to  the  power ;  consequently  in  finding  roo\Sj 
the  root  of  the  coefldcient  must  be  taken. 

The  2nd  root  of  16  a*  V  is  4  a^  b. 

Proof     Aa^b  X  ^a^b:=^  16  a'  b\ 


188  Algelra,  XXXVI. 

When  the  exponent  of  a  quantity  is  divisible  by  the  number 
expressing  the  degree  of  the  root,  the  root  can  be  found  exactly ; 
but  when  it  is  not,  the  exponent  of  the  root  will  be  a  fraction. 

The  second  root  of  a^  is  a^.     The  second  root  of  a  is  or. 
1  J. 

The  third  root  of  a  is  or.     The  nXh  root  of  a  is  a  ».      The  wth 

root  of  a™  is  an . 

The  root  of  a  fraction  is  found  by  taking  the  root  of  its  nu- 
merator and  of  its  denominator.  This  is  evident  from  the  me- 
thod of  finding  the  powers  of  fractions. 

The  root  of  any  quantity  may  be  expressed  by  enclosing  it  in 
a  parenthesis  or  drawing  a  vinculum  over  it,  and  writing  a  frac- 
tional exponent  over  it,  expressive  of  the  root.    Thus 

The  3d  root  of  8  a^  6  is  expressed 


The  root  of  a  compound  quantity  may  be  expressed  in  the 
same  way. 

The  4th  root  of  a''  +  5  a  &  is  expressed 

{a^  +  bah)^  ox  a"  +  5  a"#. 

When  a  compound  quantity  has  an  exponent,  its  root  may 
be  found  in  the  same  manner  as  that  of  a  simple  quantity. 

The  3d  root  of  (2  6  —  af  is  (2  6  —  a)^  —  (2 h—a)\ 

With  regard  to  the  signs  of  roots  it  may  be  observed,  that  all 
even  roots  must  have  the  double  sign  ±;  for  since  all  even 
powers  are  necessarily  positive,  it  is  impossible  to  tell  whether 
the  power  was  derived  from  a  positive  or  negative  root,  unless 
something  in  the  conditions  of  the  question  shows  it.  An  even 
root  of  a  negative  quantity  is  impossible.  All  odd  roots  will 
have  the  same  sign  as  the  power. 

15.  What  is  the  second  root  of  9  a^  b^  ? 

16.  What  is  the  third  root  of—  125  a*  b^  c  f 

17.  What  is  the  fifth  root  of  32  a"  a;«  r .? 


XXX VIL  Roots  of  Compound  Quantities.  189 

18.  What  is  the  third  root  of  -^1=  ? 

19.  What  is  the  fourth  root  of  %^  f 

0  m 

20.  What  is  the  second  root  of  (2  w  —  a?)®  ? 

21.  What  is  the  6th  root  of  (3  a  +  xY  ^ 

XXXVII.     Roots  of  Compound  Quantities, 

When  a  compound  quantity  is  a  perfect  power,  its  root  majc 
be  found  ;  and  when  it  is  not  a  perfect  power,  its  root  may  be 
found  by  approximation,  by  a  method  similar  to  that  employed 
for  finding  the  roots  of  numeral  quantities. 

First  we  may  observe,  that  no  quantity  consisting  of  only  two 
terms  can  be  a  complete  power ;  for  the  second  power  of  a  bi- 
nomial consists  of  three  terms  ;  that  of  a  +  ^^  f^^  example,  is 
a^  -^-^ax  -\-  x^.  The  quantity  a^  +  b^  is  not  a  complete  se- 
cond power. 

Let  it  be  required  to  find  the  second  root  of 

9x'a'  -}-4an*-\~  12  x^  a'  h\ 

The  root  of  this  will  consist  of  at  least  two  terms.  The  se- 
cond power  of  the  binomial  a  -\-  h  \s  a^  -{-  2  ah  -\-  h^.  This 
shows  that  the  quantity  must  be  arranged  according  to  the 
powers  of  some  letter  as  in  division,  for  the  second  power  of 
either  term  of  the  root  will  produce  the  highest  power  of  the 
letters  in  that  term. 

Arrange  the  above  according  to  the  powers  of  x, 

9  X*  a'  +  12  x'  a'  b'  -^4  a'  b\ 

The  formula  a^  -\-2ab  -\-  ¥  shows  that  we  should  find  the 
first  term  a  of  the  root  by  taking  the  root  of  the  first  term  ;  the 
same  must  be  the  case  in  the  given  example. 

The  root  of  9  x*  a^  is  3  x^  d\  Write  this  in  the  place  of  a 
quotient,  and  subtract  its  second  power.  Then  multiply  3  x^  a* 
by  2  for  a  divisor,  answering  to  2  a  of  the  formula. 


190  mgehra,  XXXVII. 


Divide  the  next  term  by  the  divisor.  This  gives  2ah'^  for 
the  next  term  of  the  root.  Raise  the  whole  root  then  to  the 
second  power  and  subtract  it.  Or,  which  is  the  same  thing, 
since  the  second  power  of  the  first  term  has  already  been  sub- 
tracted, write  the  quantity  2  «  ft'*  at  the  right  of  the  divisor  as 
well  as  in  the  root.  Multiply  the  whole  divisor  as  it  then  stands 
by  the  last  term  of  the  root.  This  produces  the  terms  corre- 
sponding to  2  a  6  +  &^  =  6  (2  a  -|-  6)  of  the  formula.  This  pro- 
duces 12 x^ a*¥  ^A (^ b*,  which  being  subtracted,  there  is  no 
remainder.  Consequently  the  root  is  ^x^ a^  -\-2  aV  ox  — 
^x^  a^  —  2  «  6^  The  second  power  of  both  is  the  same.  If 
the  double  sign  had  been  given  to  the  first  term  of  the  root,  the 
second  would  have  had  it  also,  and  the  positive  and  negative 
roots  would  have  been  obtained  together. 

Let  it  be  required  to  find  the  2d  root  of 

36  a=  m*  —  60  a  J  m'  4-  25  h\ 
36  a2  wi*  —  60  «  6  w'  -f  25  ¥  {Qam^-^bh 
36  a' m* 


—  60a6m'^  +  25  6^  (12am'—56 

—  60a6mV+25  6'^ 


The  process  in  this  case  is  the  same  as  in  the  last  example. 
The  second  term  of  the  root  has  the  sign  —  in  consequence  of 
the  term  QtO  ahra^  of  the  dividend  being  affected  with  that  sign. 
If  the  quantity  had  been  arranged  according  to  the  powers  of 
the  letter  6,  thus,  25  6^  —  60  a  6  m'  -f  36  a'  m\  the  root  would 
have  been  5  6  —  6  a  m^  instead  of  6  a  m^  —  5  6.  Both  roots 
are  right,  for  the  second  powers  of  the  two  quantities  are  the 


XXXVII.  Roots  of  Compound  Quantities,  lOl 

same.  The  second  power  a  —  h  is  the  same  as  that  of  6  —  a. 
One  is  the  positive  and  the  other  the  negative  root.  If  the  dou- 
ble sign  be  given  to  the  first  term  of  the  root,  both  results  will 
be  produced  at  the  same  time  in  either  arrangement. 

2ob^--'60abm^  -\-S6a^m'  {±5b^6am^ 
—  60a6m'  +  36a'^m* 


In  dividing  —  60  a  bm^  by  zh  10  b,  both  srgns  are  changed, 
the  -f-  to  — ,  and  the  —  to  +•  This  gives  to  the  second  term 
the  sign  =F .  The  first  value  is  5  b  —  6  a  m^ ,  and  the  second  is 
6am^  —  5  b. 

When  the  quantity  whose  second  root  is  to  be  found,  con- 
sists of  more  than  three  terms,  it  is  not  the  second  power  of  a 
binomial,  but  of  a  quantity  consisting  of  more  than  two  terms. 
Suppose  the  root  to  consist  of  the  three  terms  m-{-  n  -{-p.  If 
we  represent  the  two  first  terms  m  -\-nhy  I,  the  expression  be- 
comes I  -\-p,  the  second  power  of  which  is 

l2  ^2lp+p\ 

Developing  the  second  power  l^  of  the  binomial  m  -\- n,  it 
becomes  m^  -^2mn  -\-  n^.  This  shows  that  when  the  quantity 
is  arranged  according  to  the  powers  of  some  letter,  the  second 
root  of  the  first  term  will  be  the  first  term  m  of  the  root.  If  m^  be 
subtracted,  and  the  next  term  be  divided  by  2  m,  the  next  term 
n  of  the  root  will  be  obtained.  If  the  second  power  ofm-\-n 
or  l^  be  subtracted,  the  remainder  will  he2lp  -\-p^.  If  the 
next  term  2lphe  divided  by  2  Z  equal  to  twice  m  -{-  n,  the 
quotient  will  be  p,  the  third  term  of  the  root.  The  same  prin- 
ciple will  extend  to  any  number  of  terms. 

It  is  required  to  find  the  second  root  of 

4  a*  +  12  a^'  0?  +  13  a'  0?'  +  6  «  0?'  +  cc\ 
Let  this  be  disposed  according  to  the  powers  of  a  or  of  a?. 


192  ..  Algebra.  XXXVII. 

a?*  -f  6  a  a?'  +  13  tt' a?*  +  \2a^x  -f  4  a*  (a?''  +  3a^  +  2  a* root. 
^' 

1st  dividend. 

*        Qax'' -^-l^a'x^     (2a:' +  3aa?  1st  divisor. 
6  a  a;=*  +  9  a'  a?' 


2ddivid.  *  4a'a:'' -f- 12a='a? +  4a*  (2a;*  + 6aa? +2  a' 2d.  di. 
.      4a'a;''  +  12«'aJ  +  4a* 


The  process  is  so  similar  to  that  of  numeral  quantities  that 
it  needs  no  farther  explanation. 

The  double  sign  need  not  be  given  to  the  terms  during  the 
operation.  All  the  signs  may  be  changed  when  the  work  is 
done,  if  the  other  root  is  wanted.  This  will  seldom  be  the 
case  when  all  the  terms  are  positive ;  but  when  some  of  the 
terms  are  negative,  if  it  is  not  known  which  quantities  are  the 
largest,  the  negative  root  is  as  likely  to  be  found  first  as  the 
positive.  When  this  happens  the  positive  will  be  found  by 
changing  all  the  signs. 

Examples. 

1.  What  is  the  second  root  of 

4a'x'i-Ga'x'-\-a'-^  x*+4ax'? 

2.  What  is  the  second  root  of 

ifl-lf +l+^^-2^^? 

2  2  16 

3.  What  is  the  second  root  of 

—  4  a?*  +  4a;'  +  12a;'  —  6a;  +  37=^  +  9? 

4.  What  is  the  second  root  of 

x^  +  20  x'  -f  25a;*  -|-  16  +  4a:'  -j-lOa;*  +  24 a?? 


XXXVIII.  Roots  of  Compound  Quantities.  193 


XXXVIII.      Extraction  of  the  Roots  of  Compound  Quantities 
of  any  Degree. 

By  examining  the  several  powers  of  a  binomial,  and  observing 
that  the  principle  may  be  extended  to  roots  consisting  of  more 
than  two  terms,  we  may  derive  a  general  rule  for  extracting 
roots  of  any  degree  whatever, 

{a  -\-  xY  ■=.  a  -{•  X 

a  -\-x  , 


+ 

xy 

= 

a^-\-  ax 

ax  -\-  x^ 

'{a 

a''-\-2ax-\-x^ 
a  -\-  X 

+ 

xf 

= 

a'-J(-2a'x-\-ax^ 
a^  X  -\-2  ax^ 

Jr^[ 

{a 

a  -\-x 

Jrco^ 

+ 

xy 

= 

a^  a?  +  3  a' a?" 

+  aa;' 

+  x' 

(a 

;  a'  -{-Aa^x^^d'x 
a  +x 

^  -\-4:ax^ 

+  ^* 

a'  -\-4:a'x-^^a'x- 
a*a?  +  ^a^x 

+  ^^ 

(a -I-  xy  =  a'  +  5a'a;  +  10a'a)'+  \0  a""  x^ -\- ^  a  x^ -^x" 

By  examining  these  powers,  we  find  that  the  first  term  is  the 
first  term  of  the  binomial,  raised  to  the  power  to  which  the  bi- 
nomial is  raised.     The  second  term  consists  of  the  first  term 
of  the  binomial  one  degree  lower  than  in  the  first  term,  multi- 
17 


194  Algebra,  XXXVUI. 

plied  by  the  number  expressing  the  power  of  the  binomial,  and 
also  by  the  second  term  of  the  binomial.  This  will  hereafter 
be  shown  to  be  true  in  all  cases. 

The  application  will  be  most  easily  understood  by  a  particu- 
lar example. 

Let  it  be  required  to  extract  the  5th  root  of  the  quantity 

32  a'"—  80  a^  6'-|-  80  a^  6«—  40  a'  5^  +  10  a"  b''  —  b''  (2  a^  —  b^ 

32  a'' 

Dividend. 

^    ^  SO  a^b^  80  a«     divisor. 

The  quantity  being  arranged  according  to  the  powers  of  a,  I 
seek  the  fifth  root  of  the  first  term  32  a'\  It  is  2  a\  This  I 
write  in  the  place  of  the  quotient  in  division.  I  subtract  the 
fifth  power  of  2  a%  which  is  32  a'°,  from  the  whole  quantity. 
The  remainder  is 

—  80a^P-{-S0a'b'  —  &c. 

The  second  term  of  the  fifth  power  of  the  binomial  a  -\-  x 
being  5  a*  x  shows  that  if  the  second  term  in  this  case  be  di- 
vided by  five  times  the  4th  power  of  2  a^,  the  quotient  will  be 
the  next  term  of  the  root.  The  4th  power  of  2  a*  is  16  a^  and 
5  times  this  80  a^  Now  —  80  a'  b^  being  divided  by  80  a^ 
gives  —  b^  for  the  next  term  of  the  root.  Raising  2  a^  —  b^ 
to  the  fifth  power,  it  produces  the  quantity  given.  If  the  root 
contained  more  than  two  terms  it  would  be  necessary  to  sub- 
tract the  5th  power  of  2  a^  —  b^  from  the  whole  quantity  ;  and 
then  to  find  the  next  term  of  the  root,  divide  the  first  term  of 
the  remainder  by  five  times  the  4th  power  of  2  a^  —  b^.  The 
first  term  only  however  would  be  used  which  would  be  the 
same  divisor  that  was  used  the  first  time. 

When  the  number  expressing  the  root  has  divisors,  the  roots 
may  be  found  more  easily  than  to  extract  them  directly.  The 
second  root  of  a*  is  a^,  the  second  root  of  which  is  a.  Hence 
the  4th  root  may  be  found  by  two  extractions  of  the  second 
root.  The  second  root  of  a^  is  a^,  or  the  3d  root  of  a^  is  a^. 
Hence  the  6th  root  may  be  found  by  extracting  the  2d  and  3d 
roots.  The  8th  root  is  found  by  three  extractions  of  the  2d 
root,  &c. 


XXXIX.  Extraction  of  Roots.  195 

Examples, 
1.  What  is  the  3d  root  of 

'  2.  What  is  the  third  root  of 

1 5  a;*  —  6  a;  +  a;"  —  6  a;' —  20  a?' +  1 5  a;'^  +  1  f 

3.  What  is  the  4th  roc*t  of 

216  a^  x'  —  216  a  a;'  -f  81  :e*  +  16  a*  —  96  a^x? 

4.  What  is  the  5  th  root  of 

SOa;^'  — 40a?'  +  32a?'  — 80  a?*  — 1  +  10a;? 

XXXIX.     Extraction  of  the  Roots  of  Numeral  Quantities  of  any 

Degree, 

By  the  above  expression  of  the  several  powers,  we  may  ex- 
tract any  root  of  a  numeral  quantity.  Let  us  take  a  particular 
example. 

What  is  the  5th  root  of  5,443,532,400,000  ?. 

In  the  first  place  we  observe  that  the  5th  power  of  10  is 
100000,  and  the  5th  power  of  100  is  10000000000.  Therefore 
if  the  root  contains  a  figure  in  the  ten's  place,  it  must  be  sought 
among  the  figures  at  the  left  of  the  first  five  places  counting 
from  the  right.  Also  if  the  root  contains  a  figure  in  the  hun- 
dred's place,  it  must  be  sought  at  the  left  of  the  first  ten  figures. 
This  shows  that  the  number  may  be  divided  into  periods  of 
five  figures  each,  beginning  at  the  right.  The  number  so  pre- 
pared will  stand 

544,35324,00000  (340 
243 


Dividend.  3013  (405     Divisor. 

544  35324 


00000 


196  Algehra,  XXXIX. 

In  the  first  place  I  find  the  greatest  5th  power  in  544.  It  is 
243,  the  root  of  which  is  3.  I  write  3  in  the  root,  and  subtract 
243,  the  5th  power  of  3,  from  244.  The  remainder  must  con- 
tain 5  a*  ^  +  10  a^  j;'  -f-,  &c.  The  3,  that  part  of  the  root  al- 
ready found,  and  which,  by  the  number  of  periods,  must  be 
300,  answers  to  a  in  the  formula.  5  a*,  that  is,  five  times  the 
fourth  power  of  300  will  form  only  an  approximate  divisor, 
since  the  remainder  consists  of  several  terms  besides  5  a"  a; ; 
still  it  will  enable  us  to  judge  very  nearly,  and  we  shall  find 
the  right  number  after  one  or  two  trials.  As  the  fourth  power 
of  30  will  have  no  significant  figihre  below  10000,  (we  may 
consider  3  to  be  in  the  ten's  place,  with  regard  to  the  next 
figure  to  be  found,)  we  may  bring  down  only  one  figure  of  the 
next  period  to  the  remainder  for  the  dividend,  and  use  5  times 
the  fourth  power  of  3  for  the  divisor.  The  dividend  is  3013 
and  the  divisor  405.  The  dividend  contains  the  divisor  at 
least  6  times,  but  probably  6  is  too  large  for  the  root.  Try  5. 
This  gives  for  the  first  two  figures  35.  Raise  35  to  the  5th 
power  and  see  if  it  is  equal  to  544,25324.  It  will  exceed  it. 
Therefore  try  4.  The  fifth  power  of  34  is  544,35324.  Hence 
34  is  right.  Subtract  this  from  the  number,  there  is  no  re- 
mainder. There  is  still  another  period,  but  it  contains  no  sig- 
nificant figure,  therefore  the  next  figure  is  0,  and  the  root  is 
340.  The  5th  power  of  340  is  5,443,532,400,000.  If  there 
had  been  a  remainder  after  subtracting  the  5th  power  of  34,  it 
would  have  been  necessary  to  bring  down  the  next  figure  of 
the  number  to  it  to  form  a  dividend,  and  then  to  divide  it  by  5 
times  the  4th  power  of  34  ;  and  to  proceed  in  all  respects  as 
before. 

The  process  of  extracting  roots  above  the  second  is  very  te- 
dious. A  method  of  doing  it  by  logarithms  will  hereafter  be 
shown,  by  which  it  may  be  much  more  expeditiously  per- 
formed. 

Examples. 

1.  What  is  the  5th  root  of  15937022465957  .? 

2.  What  is  the  4th  root  of  36469158961  .? 

For  this,  the  fourth  root  may  be  extracted  directly,  or  it  may 
be  done  by  two  extractions  of  the  second  root.  Let  the  learner 
do  it  both  ways. 


XL.  Fractional  Exponents.  197 

3.  What  is  the  6th  root  of  481890304  ? 

This  may  be  done  by  extracting  the  6th  root  directly,  or  by 
extracting  first  the  second  and  then  the  third  root.  Let  it  be 
done  both  ways. 

4.  What  is  the  7th  root  of  13492928512  ? 

XL.     Fractional  Exponents  and  Irrational  Quantities. 

The  method  explained  above,  Art.  XXXVI,  for  extracting 
the  roots  of  Hteral  quantities,  gives  rise  to  fractional  exponents, 
when  they  cannot  be  exactly  divided  by  the  number  expressing 
the  root.  Since  quantities  of  this  kind  frequently  occur,  ma- 
thematicians have  invented  methods  of  performing  the  differ- 
ent operations  upon  them  in  the  same  manner  as  if  the  roots 
could  be  found  exactly ;  and  thus  putting  off  the  actual  ex- 
tracting of  the  root  until  the  last,  if  it  happens  to  be  most  con- 
venient. The  expressions  also  may  often  be  reduced  to  others 
much  more  simple,  and  whose  roots  may  be  more  easily  found. 

It  has  been  already  observed  that  the  root  of  a  quantity  con- 
sisting of  several  factors,  is  the  same  as  the  product  of  the 
roots  of  the  several  factors. 

H^ce        {a'  6^)*  =  (««)*.  (b')^  =  a'  b. 

(a')*  =  (a^)4.  (a)*  =  (a)*,  (a)*,  (a)* 

=  a^.  a^.  a^  =:  a.  a^  =  a  ^  =  a^. 

We  see  that  the  same  expression  may  be  written  in  a  great 
many  different  forms.     The  most  remarkable  of  the  above  are 

On  this  principle  we  may  actually  take  the  root  of  a  part  of 
the  factors  of  a  quantity  when  they  have  roots,  and  leave  the 
roots  of  the  others  to  be  taken  by  approximation  at  a  conve- 
nient time. 

The  quantity  (72  a^  b^  cy  may  be  resolved  into  factors  thug. 

^    {%yi  36  a'ab'b  c)*  =:  (36  a'  b*)^.  (2  a  b  z-^*, 
17* 


198  Algebra.  XL. 

The  root  of  the  first  factor  36  a^  h*  can  be  found  exactly, 
and  the  expression  becomes 

:  6  a  6^  (2  a  6  c)*.  '       ' 

This  expression  is  much  more  simple  than  the  other,  for  now 
it  is  necessary  to  find  the  root  of  only  2  ah  c. 
The  expression  might  have  been  put  in  this  form, 

(72)*  a^  h^  c*  =  (36.2)*  «^*  6^*  c*  =  6.2*  a  a*  h'  6*  c* 

=  6  a  h'{2  a  h  c)* 

Examples. 

1.  Reduce  (16  a'  h*y  to  its  simplest  form. 

Am,  2ab{2a'  6)*. 

2.  Reduce  (54  a  a?')*  to  its  simplest  form. 

3.  Reduce  I  _ .^^i  p  to  its  simplest  form. 

Ans.  ^-^  (l±Y. 
7  6  \3  6  c/ 

4.  Reduce  (16  a'  b'  +  32  a^b'  w)*  to  its  simplest  form. 
(16  (^  6^  -I-  32  a'  6^  m)*  =  (16  a'  6^)*   (a  6'  +  2  6  m)* 

^m.  4  a  6  (a  5'  +  2  6  m)*     ■ 

5.  Reduce  /  135^V-108a^V\ *  ^^  .^^  ^j      j^^^  ^^^^         . 

V  64wi^7i»  / 

Sometimes  it  is  convenient  to  multiply  a  root  by  another 
quantity,  or  one  root  by  another. 

If  it  is  required  to  multiply  (3  a'  6)*  by  a  6,  it  may  be  express- 
ed thus  :  ab{Sa^ by.  But  if  it  is  required  actually  to  unite 
them,  a  b  must  first  be  raised  to  the  second  power,  and  the  pro- 


XL.  Fractional  Exponents.  199 

duct  becomes  (3  a*  6^)*.     This  will  appear  more  plain  in  the 
following  manner, 

(3a^6)*=:3*  a  6* 
This  multiplied  by  a  6  is 

3*  a  6*  X  «  6  =  3*  a^  h^^  =  3*  a'  b^  =  (3  a'  b^)^. 

If  instead  of  enclosing  the  quantity  in  the  parenthesis  and 
writing  the  exponent  of  the  root  over  it,  we  divide  the  expo- 
nent of  all  the  factors  by  the  exponent  of  the  root,  all  the  ope- 
rations will  be  very  simple. 

Let  a*  be  multiplied  by  a^ . 

a^  X  a^  =  a^""  *  =  a. 

a^  X  a^  =a^^^  =a^  =za\ 

a^   b^  X  a^   b^  z^a"^*^   b^^^  z=i  a^   bK 

That  is,  multiplication  is  performed  on  similar  quantities  by 
adding  the  exponents,  as  when  the  exponents  are  whole  num- 
bers. In  like  manner  division  is  performed  by  subtracting  the 
exponents. 


5 


It  must  be  observed  that  a^  may  be  read,  the  third  root  of  the 
second  power  of  a,  or  the  second  power  of  the  third  root  of  a.     For 

a 

the  3d  root  of  a'  is  a^ ;  and 

X  i  i    I    i  2 

a^  X  a^  =a^~^  =a3. 


The  3d  power  of  a^  is 

'  X  a^  Xa'^  —a^  '  «'  ^  ==««'- ^  ==  « 


2  2  _L.2     I     2  2  V  f?  « 


That  is,  a  power   of  a  root  may  be  found  by  multiply- 
ing the  fractional  exponent  by  the  exponent  of  the  power. 


260  Algebra.  XL. 

Consequently  a  root  of  a  root  may  be  found  by  dividing  the 
fractional  index  by  the  exponent  of  the  root.  In  multiplying 
and  dividing  the  fractional  exponents,  we  must  apply  the.  same 
rules  that  we  apply  to  common  fractions. 

3  i 

The  3d  root  of  a ^  is  a*. 
The  3d  root  of  a^  is  J. 
The  5th  root  of  a^   Jt  jg  ^t^  520, 

If  the  numerator  and  denominator  both  be  multiplied  or  di- 
vided by  the  same  number,  the  value  of  the  quantity  will  iicJt 
be  altered  ;  for  that  is  the  same  as  raising  it  to  a  power,  and 
then  extracting  the  root. 

a.  4  10 

2  i 

If  it  is  required  ta  multiply  «»  by«^,  the  fractions  may  be 
reduced  to  a  common  denominator  and  added  :  thus, 

2.  X  A  1  1  \i  i 

a*  X  a    =  a^  X  «®  =  «^  =  a  ^  =  a   a®. 

The  same  may  be  done  in  division  and  the  exponents  sub- 
tracted. 

7  2_i 

a  2  r,    6  2   1     _    10  II 

—  —  /»  o  0     —T  n  8 


.  a 


5.  10 

3  «   6 


9     __  2  tf  __   1 


^  ri  !1^  S5  a^^       TT~  a  —  T¥  -2 ,i^. 

5  2^  IJL 

In  fact,  quantities  with  fractional  exponents  are  subject  to 
precisely  the  same  rules,  as  when  the  exponents  are  whole 
numbers ;  but  the  rules  must  be  applied  as  to  fractions.  The 
fractions  may  be  reduced  to  decimals  without  altering  the 
value  ;  thus 

i         li         1.25  .25  -2         .06 

a*  =  fl  *  =  a         —ay,  a       =z  a  X  a     X  » 


XL.  Fractional  Exponents,  201 

«J  X  «^  =  «'•"'  X  a^  =  a2-35  =  a^xJ'-X  a^^^' 

It  is  very  important  to  remember  how  these  quantities  may- 
be separated  into  factors.  Since  multiplication  is  performed 
by  adding  the  exponents,  and  division  by  subtracting  them,  any 
quantity  may  be  separated  into  as  many  factors  as  we  please, 
by  separating  the  exponent  into  parts.     Thus, 

a^  =  a^  X  a^  =  a  X  a"^  =  aX  a^  X  a"" 

=:  a^  X  a^  X  «'  X  «^  X  «^  X  «^  X  a^. 

The  sum  of  all  the  exponents  in  the  last  expression  is  5.  Lo- 
garithms are  of  the  same  nature  as  these  exponents,  and  afford 
as  great  a  facility  in  operating  upon  numbers,  as  these  do  upon 
letters.  And  the  operations  are  performed  in  the  same  way, 
as  will  be  explained  hereafter. 

If  the  learner  should  ever  have  occasion  to  read  other  trea- 
tises on  mathematics,  he  will  generally  find  the  roots  express- 
ed by  what  are  called  radical  signs.     The  second  root  is   ex- 

.  3    

pressed  with  the  sign  \/     ,  the  third  root  \/      the  same  sign 

4 

with  the  index  of  the  root  over  it.     The  4th  root  is  \/     ,  (fee. 


J 

=  ^V 

«* 

=   V^ 

ai 

=  VT 

3 

=  V"" 

2* 

2          5 
«3      6^- 

^2d 

'h%. 

(fee. 

They  will  be  easily  understood  if  the  radical  sign  be  removed, 
and  the  exponents  divided  by  the  index  of  the  root  or  the  quan- 
tity enclosed  in  a  parenthesis,  and  the  root  written  over  it. 

4  

The  expression  \/  id  a^h^  becomes 

5*  a^  h^  =  (5  a'  h')^. 
The  expression  ^y/  a*  -f-  6"  is  equivalent  to  (a^  -j-  b^)^ 


202  Algebra,  XLI. 

XLI.     Binomial  Theorem. 

It  has  already  been  remarked  that  the  powers  of  any  tjuanti- 
ty  are  found  by  multiplying  the  quantity  into  itself  as  many 
times,  less  one,  as  is  expressed  by  the  exponent  of  the  power. 
Sir  Isaac  Newton  discovered  a  method,  by  which  any  quantity 
consisting  of  more  than  one  term  may  be  raised  to  any  power 
whatever,  without  going  through  the  process  of  multiplication. 

The  principle  on  which  this  method  is  founded  is  called  the 
Binomial  Theorem.  Its  use  is  very  important  and  extensive  in 
algebraic  operations. 

Next  to  quantities  consisting  of  only  one  term,  binomials,  or 
quantities  consisting  of  two  terms,  are  the  most  simple. 

Let  a  few  of  the  powers  of  a  +  a?  be  found  and  their  forma- 
tion attended  to. 


{a  +  xy  = 

a-\-x     jt 

d>-\-ax, 

ax  -\-  oc^ 

(a  +  xY  = 

a"  +  2ax-\-x^ 

a^  '{-^a'x-^-ax^ 

a'x+^ax^  +x^ 

{a  +  xY  = 

a^-{-^a^x  +  '^a^x^-\-ax^ 

(a  +  xy  = 

a'  +  A  a^  X  -\-  ^  a"  x^  +  Aax^  -\-  x' 
a-\-  X 

• 

a'j^  A  a'  X  +  ^  a^  x^  -\-  A  a" x^  -\-  ax" 

a^x-\-Aa^x^'^Qa^x^  -f4«a:*  +a?' 

(a  +  a?)*=:    G'4-5a4a:+  lOa^a?'.}.  10 a' a?^  J^^ax^  -\-x\ 


XLI*  Binomial  Theorem,  203 

The  law  of  the  formation  of  the  literal  part  is  sufficiently 
manifest. 

In  each  power  there  is  one  term  more  than  the  number  de- 
noting the  power  to  which  it  is  raised.  The  first  power  con- 
sists of  two  ternis,  the  second  power  of  three  terms,  the  third 
power  of  four  terms,  &c. 

In  every  power  a  is  found  in  every  term  except  the  last,  and 
X  is  found  in  every  term  except  the  first.  The  exponent  of  a  in 
the  first  term  is  the  same  as  the  exponent  of  the  power  to 
which  the  binomial  is  raised,  and  it  diminishes  by  one  in  each 
succeeding  term. 

The  exponent  of  a?  in  the  second  term  is  1,  and  it  increases 
by  one  in  each  succeeding  term,  until  in  the  last  term  it  is  the 
same  as  that  of  a  in  the  first  term. 

The  law  of  the  coefficients  is  not  so  simple,  though  it  is  not 
less  remarkable. 

The  coefficients  of  the  first  power,  viz.  a  -\-  x,  are  1,1;  those 
of  the  second  power  are  1,  2,  1.  These  are  formed  from  the 
first  as  follows.  When  a  is  multiplied  by  a,  it  produces  a*, 
and  no  other  term  being  produced  like  it,  there  is  nothing  add- 
ed to  it,  and  it  remains  with  the  same  coefficient  as  the  a  in  the 
multiplicand.  In  multiplying  x  by  a  and  afterward  a  by  x, 
two  similar  terms  are  produced,  having  the  coefficients  of  the 
a  and  x  in  the  multiplicand,  viz.  1  and  1  ;  and  the  addition  of 
these  forms  the  2.     The  other  1  is  produced  like  the  first. 

The  coefficients  of  the  third  power  are  1,  3,  3,  1.  The  Is 
are  produced  from  the  second  power,  as  those  of  the  second 
power  are  produced  from  the  first.  In  multiplying  2  axhy  a, 
the  term  produced  is  2  a^  x,  having  the  coefficient  of  the  se- 
cond term  of  the  multiplicand  ;  and  in  multiplying  a^  by  x,  the 
term  produced  is  a^  x,  similar  to  the  last,  and  having  the  coeffi- 
cient 1  of  the  first  term  of  the  multiplicand.  The  addition  of 
the  coefficients  of  these  two  terms  produces  the  3  before  a^  x. 
That  is,  the  coefficient  of  the  second  term  of  the  third  power  is 
formed  by  adding  together  the  coefficients  of  the  first  and  se- 
cond terms  of  the  second  power.  In  the  same  manner  it  may 
be  shown,  that  the  coefficient  3  of  the  third  term  of  the  third 
power  is  formed  by  adding  together  the  coefficients  of  the  se- 
cond and  third  terms  of  the  second  power. 

The  following  law  will  be  found  on  examination  to  be  ge- 
neral. 


^04  Algebra,  XLI. 

The  coefficient  of  the  first  term  of  every  power  is  1.  The 
coefficient  of  the  second  term  of  every  power  is  formed  by  add* 
ing  together  the  coefl[icients  of  the  first  and  second  terms  of 
the  preceding  power.  The  coefficient  of  the  third  term  of 
every  power  is  formed  by  adding  together  the  coefficients  of 
the  second  and  third  terms  of  the  preceding  power.  The  co- 
efficient of  the  fourth  term  of  every  power  is  found  by  adding 
together  the  coefficients  of  the  third  and  fourth  terms  of  the 
preceding  power.     And  so  of  the  rest. 

This  law,  though  perhaps  sufficiently  evident  by  inspection, 
may  be  easily  demonstrated. 

Suppose  the  above  law  to  hold  true  as  far  as  some  power 
which  we  may  designate  by  n»  The  literal  part  of  the  nth 
power  will  be  formed  thus. 

a»,  a^-'  .r,  a""'  x%  a"""^  x^ a  oc^-\  x\ 

We  cannot  write  all  the  terms  without  assigning  a  particular 
value  to  n.  We  can  write  a  few  of  the  first  and  last.  The 
points  between  show  that  the  number  of  terms  is  indeterminate  ; 
there  may  or  may  not  be  more  than  are  written. 

Suppose  that  A  is  the  coefficient  of  the  second  term,  B  that 
of  the  third,  &c.  and  let  the  whole  be  multiplied  by  a  +  x, 
which  will  produce  the  next  higher  power,  or  the  (w  -{-  l)th 
power. 


XLI. 


Binomial  Ttieorem* 


205 


t 

+ 

r 

+ 
+ 

'- 

ST 

t 

I 

1^ 

• 

«  + 

% 

1  '^ 

to 

+ 

T 

+ 

I 

k    : 

O 

f^ 

«    . 

C3 

"e    • 

+ 

U, 

fe  1, 

:  1 

L 

!    s. 

« 

*    ?s 

.*  O 

O 

+ 

+  4- 

+ 

m 

w       eo 

PQ 

T 

8     Sj 

T    T 

+ 

t 

^«  ^« 

"?i 

o 

o  m 

I 

+ 

+  + 

« 

-*« 
? 

+ 

L 

< 

« 

e     ^ 

« 

w  < 

+ 

+ 

+  + 

^^ 

?s 

< 

I 

< 

+ 

<  ^ 

+ 

Hh 

+  + 

1 

1 

I 

18 


206  Algebra.  XLI. 

In  this  result  we  observe  that  the  exponents  of  both  a  and  x 
are  increased  by  1  in  each  term,  and  there  is  still  one  term 
without  X  and  another  without  a.  Before  the  terms  of  the  pro- 
duct were  added,  there  were  twice  as  many  terms  in  the  pro- 
duct as  in  the  multiplicand,  but  they  have  all  united  two  by 
two  except  the  first  and  last.  The  terms  C  «^~"^  x'^  and  Fa* 
x'^~^  have  not  united  with  any  others,  but  it  is  evident  that  they 
would  have  done  so,  if  all  the  terms  could  have  been  written. 
There  is  then  one  more  term  in  this  power  than  in  the  last. 

The  coeflicient  of  the  first  term  is  still  1 .  That  of  the  se- 
cond is  the  sum  of  the  coefficients  of  the  first  and  second  terms 
of  the  multiplicand,  viz.  1  +  A.  That  of  the  third  is  the  sum 
of  the  coefficients  of  the  second  and  third  terms  of  the  multi- 
plicand, viz.  A  -|-  B  ;  &c. 

The  above  formula  shows  that  if  the  law  above  mentioned  is 
true  for  one  power,  it  will  be  so  for  the  next  higher  power. 
We  have  seen  that  it  is  true  for  the  5th  power,  therefore  it  will 
be  true  for  the  6th  ;  being  true  for  the  6th,  it  will  be  so  for  the 
7th,  &C. 

Let  the  coefficients  of  several  of  the  first  powers  be  written 
without  the  letters,  forming  them  by  the  above  principle. 

First  observe  that  (a  -}-  a?)°  =  1. 

Adding  0  to  this  1  gives  1,  and  then  0  again  on  the  other 
side  gives  1.  Hence  we  have  1,  1  for  the  coefficients  of  the 
first  power. 

Adding  0  to  the  first  1  gives  1  ;  adding  1  and  1  gives  2,  and 
then  1  and  0  are  1.  Hence  the  coefficients  of  the  second  pow- 
er are  1,2,  1 . 

Again,  0  +  1  =  1;  1  -}-2=3;2  +  lr=3;  1+0=1. 
Hence  1,  3,  3,  1  are  the  coefficients  of  the  third  power. 

Again,  0  +  l  =  l;l  +  3  =  4;3  +  3  =  6;3  +  l  =  4; 
and  1  +  0  =  1.  Hence  1,  4,  6,  4,  1  are  the  coefficients  of  the 
fourth  power. 

Again,  0+l=l;4+4=:5;4+6=:10;6  +  4=:10; 
4  +  1  =  5;  and  1  +  0=1.  Hence  1,  5, 10, 10,  5,  I  are  the 
coefficients  of  the  5th  power,  &c. 


XLI.  Binomial  Theorem.  20*^ 


The  Coefficients  of  the  first  Ten  Powers, 

1 

1         1 

12  1 

1         3         3         1 

14         6         4         1 

1         5       10       10         5         1 

1         6       15       20       15         6         1 

1         7       21       35       35       21         7         1 

1         8       28       56       70       56       28         8         1 
1         9       36       84     126     126       84       36         9         1 
1        10       45     120     210     252     210     120       45        10  1 

Here  we  observe  that  the  first  row  of  figures  taken  obliquely 
downward  is  the  series  of  numbers  1,  1,  1,  &c. 

The  second  row  is  the  series  of  natural  numbers,  1,  2,  3,  4, 
5,  &c.  whose  differences  are  1. 

The  third  row  is  the  series  1,  3,  6,  10,  15,  <fcc.  whose  differ- 
ences are  the  last  series,  viz.  1,  2,  3, 4,  &c. 

The  fourth  row  is  the  series  1,4,  10,  20,  35,  <fcc.  whose  dif- 
ferences are  the  last  series,  viz.  1,  3,  6,  10,  &c.  Each  succes- 
sive row  is  a  series,  whose  differences  form  the  preceding  row. 

We  may  observe  farther  that  the  coeflfiicient  of  the  second 
term  of  any  power  is  the  term  of  the  series  1,  2,  3,  4,  &c.  de- 
noted by  the  exponent  of  the  power.  That  of  the  second  pow- 
er, is  the  second  term  ;  that  of  the  third  power,  the  third  term ; 
that  of  the  nth  power,  the  nth  term.  But  this  being  the  series 
of  natural  numbers,  the  number  which  denotes  the  place  of  the 
term  is  equal  to  the  term  itself,  so  that  the  coefficient  of  the 
second  term  will  always  be  equal  to  the  exponent  of  the 
power. 

The  coefficient  of  the  third  term  of  any  power  is  the  term  of 
the  series  1,  3,  6,  10,  &c.  denoted  by  the  exponent  of  the  pow- 
er diminished  by  1 .     That  of  the  third  power  is  the  second 


208  mgebra,  XLIL 

term,  that  of  the  fourth  power  the  third  term,  that  of  the  nth 
power  the  {n —  l)th  term,  <fcc. 

The  coefficient  of  the  fourth  term  of  any  power  is  the  term 
of  the  series  1,  4,  10,  20,  &c.  denoted  by  the  exponent  of  the 
power  diminished  by  2.  That  of  the  fourth  power  is  the  se- 
cond term,  that  of  the  fifth  power  is  the  third  term,  that  of  the 
nth  power  is  the  (w  —  2)th  term.  And  so  on  as  we  proceed  to 
the  right,  the  place  of  the  term  in  the  series  is  diminished 
by  1. 

We  may  observe  another  remarkable  fact,  the  reason  of  which 
will  be  manifest  on  recurring  to  the  formation  of  these  series. 
We  shall  take  the  7th  power  for  an  example,  though  it  is  equal- 
ly true  of  any  other. 

The  coefficient  of  the  second  term,  viz.  7,  is  the  sum  of  7 
terms  of  the  preceding  series  1,  1,  1,  &,c.  and  was  in  fact  form- 
ed by  adding  them. 

The  coefficient  of  the  third  term,  21,  is  the  sum  of  the  first 
six  terms  of  the  preceding  series,  1,  2,  3,  &c.  and  was  actually 
formed  by  adding  them,  as  may  be  seen  by  referring  to  the. for- 
mation. 

The  coefficient  of  the  fourth  term,  35,  is  the  sum  of  the  first 
five  terms  of  the  preceding  series,  1,  3,  6,  10,  &,c.  and  was 
formed  by  adding  them. 

The  same  law  continues  through  the  whole.  If  now  we  can 
discover  a  simple  method  of  finding  the  sums  of  these  series 
without  actually  forming  the  series  themselves,  it  will  be  easy 
to  find  the  coefficients  of  any  power  without  forming  the  pre- 
ceding powers.     This  will  be  our  next  inquiry. 

XLII.     Summation  of  Series  by  Differences. 

It  is  not  my  purpose  at  present  to  enter  very  minutely  into 
the  theory  of  series.  I  shall  examine  only  a  few  of  the  most 
simple  of  them,  and  those  principally  with  a  view  of  demon- 
strating the  binomial  theorem. 

A  series  by  differences  is  several  numbers  arranged  together, 
the  successive  terms  of  which  differ  from  each  other  by  some 
regular  law. 


XLII.  Summation  of  Series  by  Differences.  209 

I  call  a  series  of  the  first  order  that,  in  which  all  the  terms 
are  alike,  as  1,  1,  1,  1,  &c.  3,  3,  3,  3,  &c.  a,  a,  a,  a,  (fee.  In 
these  the  difference  is  zero. 

The  sum  of  all  the  terms  of  such  a  series  is  evidently  found 
by  multiplying  one  of  the  terms  by  the  number  of  terms  in  the 
series.  Every  case  of  multiplication  is  an  example  of  finding 
the  sum  of  such  a  series. 

The  sum  *  of  a  number  n  of  terms  of  any  series  a,  a,  a,  &c. 
is  expressed 

_  na 
s  —  ~ 

When  a  =  1,  it  becomes  s  =-, 

1 

A  series  in  which  the  terms  increase  or  diminish  by  a  con- 
stant difference,  is  called  a  series  of  the  second  order.  As  1,2,, 
3,  4,  5,  &c.  3,  6,  9,  12,  &c.  or  12,  9,  6,  3.  A  series  of  this 
kind  is  formed  from  a  series  of  the  first  order.  The  differences 
between  the  successive  terms  form  the  series  from  which  it  is 
derived. 

At  present  I  shall  examine  only  the  series  of  natural  num- 
bers 1,  2,  3,  4, n. 

This  series  is  formed  as  follows : 

0  +  1  =  1 

1  -f-  1  =2 

1+1+1=3 
1+1+1+1=4 
1  +  1  +  1  +  1  +  1  =  5,  &c. 

The  sum  of  any  number  n  of  terms  of  the  series  1,1,1,1,  &c. 
is  equal  to  the  nth  term  of  the  series  1,  2,  3,  4,  &c. 

Write  down  two  of  these  series  as  follows  and  add  the  cor- 
responding terms  of  the  two  together. 

1,     2,     3,     4,     5 

5,  4,     3,     2,     1 

6,  6,     6,     6,     6 

18* 


210  mgehra.  XLIL 

1,         2,         3,         4,  . . .  (71—3),  (n— 2),  (n— 1,)  n 
»,  (w— l),(n— 2),  (n— -3) 4,  3,  2,         1 . 

(/i+l),(n+l),(n+l),(n+l)...(n+l),(n+l),(7i+l),(w+l) 

The  6th  term  of  the  series  is  6,  and  it  appears  that  5  times 
6  will  be  twice  the  sum  of  5  terms  of  the  series. 

The  {n  -f  l)th  term  of  the  series  1,  2,  3,  4,  <fec.  is  w  +  1.  It 
appears  that  n  times  [n  •\-\)  will  be  twice  the  sum  of  n  terms  of 
the  series. 

The  sum  s'  of  any  number  n  of  terms  may  be  expressed 
thus. 


S'  Z=Z 


(71+1 


1.2 


It  is  frequently  convenient  to  use  the  same  letter  in  similar 
situations  to  express  different  values.  In  order  to  distinguish 
it  in  different  places,  it  may  be  marked  thus,  s,  s',  s",  s"\  which 
may  be  read  s,  s  prime,  s  second,  s  third,  &c. 

How  many  times  does  the  hammer  of  a  clock  strike  in  12 
hours  ? 

In  this  example  n  =  12        n  -|-  1  =  13. 

^^  ^   ^^  =  78.  Arts.  78  times. 

1  X  2 

The  rule  expressed  in  words  is  ;  To  find  the  sum  of  any  num- 
ber of  terms  of  the  series  1,2,  3,  4,  &c.  find  the  next  stu^ceeding 
term  in  the  series,  and  multiply  it  by  the  number  of  terms  in  the 
series,  and  divide  the  product  by  2. 

The  same  thing  may  be  proved  in  another  form  which  is 
more  conformable  to  the  method  that  will  be  used  for  the  series 
of  the  higher  orders. 

Suppose  it  is  required  to  find  the  sum  of  the  first  five  terms 
of  the  series. 

The  sixth  term  of  the  series  is  the  sum  of  6  terms  of  the  se* 
ries,  1,  1,  1,  &c.  thus 

l+l-|-l-hl  +  l  +  l=6. 


XLII. 


Summation  of  Series  by  Differences, 


211 


Let  this  series  be  written  down  five  times,  one  under  the 
other,  thus. 


If  this  series  be  divided  by  a  line  passing  diagonally  through 
it,  so  that  the  part  below  and  at  the  left  of  the  line  may  con- 
tain one  term  of  the  first  series,  two  of  the  second,  three  of  the 
third,  four  of  the  fourth,  and  five  of  the  fifth  ;  the  terms  so  se- 
parated will  form  the  first  five  terms  of  the  series  1,  2,  3,  &c. 
There  will  be  the  same  number  of  terms  above  and  at  the  right 
of  the  line,  which  will  form  the  same  series,  if  the  terms  be 
added  vertically  instead  of  horizontally. 


1,     1,\1,     1,     1,     1 


1,     1,     1,\1,     1,     1 


1,     1,     1,     1 
1,     I,     1,     1 


1,M 


It  is  easy  to  see  that  this  series  continued  to  any  number  of 
terms  will  be  formed  twice  over  in  this  way,  if  the  number  of 
series  written  under  each  other  is  equal  to  the  number  of  terms 
required  and  the  number  of  terms  in  each  series  exceed  the 
number  of  terms  by  one.  And  the  reason  of  it  is  manifest  from 
the  manner  in  which  the  two  series  are  formed. 

Hence  n  times  the  series  consisting  of  n  +  1  terms  of  the 
series  1,  1,  1,  1,  &c.  will  be  twice  the  sum  *'  of  w  terms  of  the 
series  1,  2,  3,  4,  &,c. 

That  is,  2  y  =  w  (n  +  1)  and  5'  =  w_(7i_-f_l;^ 


212  Algebra.  XLII. 

A  series  of  the  third  order  is  one,  the  difference  of  the  succes- 
sive terms  of  which  is  a  series  of  the  second  order.  I  shall 
consider  only  the  series  formed  from  the  series  1,  2,  3,  &c. 

Formation. 


0-f-  1  =    O-fl  =    1 

1  +2  =    1  +  2=    3 

1  +2-1-3  =    3  +  3=    6 

1+2  +  3  +  4  =:    6+4  =  10 

1  +2  +  3  +  4  +  5       =10+5=16 
1  +2  +  3  +  4  +  5  +  6=15  +  6  =  21,&c. 

The  first  term  of  the  series  1,  2,  3,  &c.  forms  the  first  term  ; 
the  sum  of  the  first  two  terms  forms  the  second ;  the  sum  of 
the  first  three  forms  the  third  term,  &c.  and  the  sum  of  w  terms 
will  form  the  nth  term  of  the  series  1,  3,  6,  10,  &c. 

Let  it  be  required  to  find  the  sum  of  the  first  five  terms  of 
the  series  1,  3,  6, 10,  15,  21,  &c. 

The  sixth  term  of  this  series  is  the  sum  of  the  first  6  terms 
of  the  series  1,  2,  3,  &c. 

1-1-2  +  3+4  +  5+6  =  ?-2^2  =  21  =  6th  terni. 

Write  this  series  five  times  one  under  the  other,  end  draw  a 
line  diagonally  so  as  to  leave  on  the  left  and  below,  the  first 
term  of  the  first,  the  first  two  of  the  second,  the  first  three  of 
the  third,  &,c.  and  the  first  five  of  the  fifth. 

1,\2,  3,  4,  5,  6 
1,  2\3,  4,  5,  6 
1,  2,  3>s^,  5,  6 
1,  2,  3,  4^\5,  6 
1,     2,     3,     4,     6\6 


XLII.  Summation  of  Series  by  Differences,  213 

The  figures  so  cut  off  form  the  first  five  terms  of  the  series 

1,  3,  6,  10,  15,  &c.  the  sum  of  which  we  wish  to  find.  It  will 
now  be  shown  that  the  sum  of  the  terms  on  the  right  and 
above  the  line,  is  equal  to  twice  the  sum  of  those  below  and 
at  the  left. 

By  the  rule  given  above  for  finding  the  sum  of  the  series  1, 

2,  3,  &c. 

_  1  X  2 


The  sum  of  1  term,  or  1 
The  sum  of  2  terms,  or  1  +  2 
The  sum  of  3  terms,  or  1  +  2  +  3 


2 

2X3 

2 

3X4 

2 


The  sum  of  4  terms,  orl-{-24-3-|-4  ==  121^. 

The  sum  of  5  terms,  or  1  +  2  +  3  +  44-5       =  £21.^. 

Hence     2  (1)  =1X2 

2  (1  +  2)  =2x3 

2(1+2  +  3)  =3X4 

2(1+2+3  +  4)  =4X5 

2(1+2  +  3  +  4  +  5)  =5X6 

That  is,  the  2  is  twice  the   1, 

The  two  threes  are  twice  (1+2), 

The  three  fours  are  twice  (1  +2  +  3), 

The  four  fives  are  twice    (1  +2  +  3  +  4),  and 

The  five  sixes  are  twice     (1  +2  +  3  +  4  +  5). 

Since  the  part  below  the  line  forms  the  series  whose  sum  is 
required,  and  the  part  above  the  line  is  equal  to  twice  that  be- 
low, both  parts  together  are  equal  to  three  times  the  series  1, 
3,  6,  10,  15.  Therefore  if  21,  which  is  the  next  term  in  the 
series,  and  which  is  also  the  sum  of  the  series  1,  2,  3,  4,  5,  6, 
be  multiplied  by  5,  the  number  of  terms  to  be  summed,  and 


211  Algebra.  XLII. 

divided  by  3,  the  quotient  will  be  the  sum  of  the  series  re- 
quired. 

It  is  easy  to  see  that  if  the  series  1,2,  3, . . .  (w  -f  1)  be  writ- 
ten n  times  and  divided  by  a  line  like  the  above,  the  part  be- 
,  low  the  line  will  form  n  terms  of  the  series  1,  3,  6,   10,  &c. 
And  the  part  above  the  line  will  be  equal  to  twice  the  part  be- 
low, because  the  sum  of  w  terms  of  the  series  1,  2,  3,  &c.  is 

1X2 

Therefore  to  find  the  sum  of  n  terms  of  the  series  1,  3,  6, 10, 
multiply  the  (n  -[-  l)th  term  of  that  series  by  n  and  divide  by 
3,  and  the  quotient  will  be  the  sum  required. 

But  the  {n  +  l)th  term  of  the  series  is  equal  to  the  sum  of 
(w-f  1)  terms  of  the  series  1,  2,  3,  4,  &c.     The  nth  term  of 

this  series  being  ^A!LjI-J,  the  {n  +  l)th  term  will  be 
1  X  '^ 

(ll+l)(7l+2) 

1     X      2       * 

This  being  multiplied  by  n,  the  number  of  terms,  and  divided 
by  3,  gives 

w(?i+l)(yi-f  2) 
1  X  2     X      3       * 

Hence  the  sum  5"  of  n  terms  of  the  series  will  be  expressed 
thus, 

1X2X3 

A  series  of  the  fourth  order  is  one,  the  difference  of  whose 
terms  is  a  series  of  the  third  order. 

I  shall  at  present  consider  only  the  one  formed  from  the 
series  1,  3,  6,  10,  15,  &c. 


XLII.  Summation  of  Series  by  Differences.  216 

Formation. 

0+1  =0+1=1 

1+3  =1+3=4 

1+3  +  6  =    4+    6=  10 

1  +  3  +  6+10  =  10  +  10  =  20 

1+3  +  6+10  +  15  =  20  +  15  =  35 

1+3+6  +  10  +  15+21  =35+21=56 

The  first  term  of  the  series  1,  3,  6,  &c.  is  the  first  term  of  the 
new  series  ;  the  sum  of  the  first  two  terms  forms  the  second  ; 
>&.c.  the  sum  of  n  terms  will  form  the  nth  term  of  the  new 
series. 

It  is  required  to  find  the  sum  of  five  terms  of  this  series. 

The  sixth  term  of  this  series  is  equal  to  the  sum  of  the  first 
six  terms  of  the  preceding. 

1  +  3  +  6  +  10  +  15  +  21  =  I>i-IA_?.  =  56. 
^^^       ^       ^  1X2X3 

Write  this  series  five  times,  one  under  the  other,  and  sepa- 
rate it  into  two  parts  by  a  line  drawn  diagonally  in  the  same 
manner  as  was  done  with  the  last  series.  The  terms  below  the 
line  will  form  the  series  whose  sum  is  required,  and  the  terms 
above  the  line  will  be  equal  to  three  times  those  below.  That 
is,  the  whole  will  be  four  times  the  sum  required. 

1,\  3,  6,     10,     15,     21 

1,  3\  6,     10,     15,     21      - 

1,  3,  6\  10,     15,     21 

1,-  3,  6,     loXl5,     21 

1,  3,  6,     10,      15^  21 


316  Algebra.  XLII. 

By  the  rule  given  above  for  finding  the  sum  of  the  series  1, 
3,  6,  10,  &c. 

The  sum  of  one  term,  or  1  -—-     =1. 

3 

\  2  V  6 

The  sum  of  two  terms,  or  1  +  3  =    —1-  =  4. 

The  sum  of  three  terms,  or  1  +  3  +  6         =  i21i?  —  10. 

3 

The  sum  of  four  terms,  or  1  +  3  +  6+10=  i^U^  =  20. 

The  sum  of  five  terms,  or  1+3+6+10+15  =  ^^^  =  35. 

The  five  21s  are  3  times  1+3  +  6  +  10  +  15. 

The  four  15s  are  3  times  1+3  +  6  +  10 

and  so  of  the  rest. 

It  is  easy  to  see  that  this  principle  will  extend  to  any  number 
of  terms. 

Therefore  to  find  the  sum  of  n  terms  of  the  series  1,  4,  10, 
20,  (fee.,  multiply  the  (n  +  l)th  term  of  the  series  by  w,  and 
divide  the  product  by  4,  and  the  quotient  will  be  the  sum  re- 
quired. 

But  the  [n  +  l)th  term  of  this  series  is  equal  to  the  sum  of 
(?i+  1)  terms  of  the  preceding  series. 

The  wth  term  of  the  preceding  series  being 

yt  (n  +  1)  (?z  +  2) 
1X2X3       ' 

the  (n  +  l)th  term  will  be 

»  (n  +  l)(n+2)(n  +  3) 

1X2X3 
This  being  multiplied  by  n  and  divided  by  4,  gives 

^n  ^  n(7l+l)(n  +  2)(7l  +  3) 

1X2X3X4 


XLIII.  Summation  of  Series  by  Differe 


XLIII.      The   principle   of  summing  these  series  may  be 
proved  generally  as  follows : 

Let  I,  a,h,  c,  d Z  be  a  series  of  any  order,  such  that 

the  sum  of »  terms  may  be  found  by  multiplying  the  {n  -{-  l)th 
term  by  n,  and  dividing  the  product  by  m.  If  I  is  the  (n  -\-  1) 
th  term,  and  s  the  sum  of  all  the  terms,  we  shall  have  by  hy- 
pothesis 

s  =■  — ,  and  m  s  =  nl, 
m 

That  is,  n  I  will  be  m  times  the  sum  of  the  series.     The  next 
higher  series  will  be  formed  from  this  as  follows : 


I      .      .      .      . 

. 

=  1st  term. 

l+a     .         .         .         . 

. 

=  2d      " 

1  +  a  +  6 

. 

=  3d      " 

1  -f-a  +  6  +  c       .       . 

. 

=  4th     " 

1  -f  a -f  6  +  c  +  <; 

=  5th     " 

l+a  +  6  +  c  +  tZ  +  .. 

..k 

=  nth     " 

1  +a-l-6-f.c  +  (Z  +  .. 

..k 

+  Z=:(7l+l)th. 

The  first  term  1  of  the  original  series  !,«,&,  &c.,  forms  the 
first  term  of  the  new  series  ;  the  sum  of  the  first  two  forms  the 
second  term ;  the  sum  of  the  first  three  forms  the  third  term, 
&c.,  and  the  sum  of  (n  +  1)  terms  forms  the  (n  -}-  l)th  term. 

Let  the  series  forming  the  {n  +  l)th  term,  be  written  n 
times,  one  under  the  other,  term  for  term.  And  let  a  line  be 
drawn  diagonally,  so  that  the  first  term  of  the  first  row,  the 
first  two  of  the  second  row,  and  n  terms  of  the  nth  row  may  be 
at  the  left,  and  below  the  line. 
19 


218 


Algebra, 
1,   \  a,     6,     c,     d,      .      A:,     / 


XLIII. 


1,     a,  \  h,     c,     d,      .      k,     I 


1,     a,      h,\c,     d,      .      k,     I 


1,     a,     h,     c,  \d,     .      kf     I 


1,     «,     J,     c,      d,\'      k,     I 


1,     a,     i,     c,     df,      .  \  yfc,     ^ 


1,     a,     b,     c,      d,      .      k,\  I 


The  terms  below  and  at  the  left  of  the  line,  form  n  terms  of 
the  new  series.  It  is  now  to  be  shown  that  the  terms  above, 
and  at  the  right  of  the  line,  are  equal  to  m  times  those  below, 
and,  consequently,  that  the  whole  together  are  equal  to  ;»  -f- 
1  times  n  terms  of  the  new  series. 


By  the  hypothesis 

The  sum  of  one  term,  or  1 

The  sum  of  two  terms,  or  1  +  a 

The  sum  of  three  terms,  or  1  +  a  +  6 

The  sum  of  four  terms,  orl+a  +  6-|-c 


=  I— 

m 

_2_b_ 
m 

=  ?-i 
m 

_Ad_ 

m 

nl 
The- sum  of  n  terms,  orl+«+5+c-j-<?  +  ..A;  = 


XLIII.  Summation  of  Series  by  Differences,  219 

Multiplying  both  members  of  the  above  equations  by  m  : 

m.  1  '  =  1   a 

m{l  -\-  a)  =2h 

m{i  +a  +  b)  =  3  c 

m(l-|-a-f6-fc)  =4d 

m{l-{-a-\-b  +  c  +  d  +  .,,k)  z=inl 

Hence  it  appears,  that  a  is  »i  times  1 ;  2  6  is  »i  times  (1  -(-  a) 

&c. ;  and  n  I  is  m  times  (1  -\-a-{-b-\-c-\-d-\- k);  that 

is,  the  part  above  and  at  the  right  of  the  line,  is  m  times  the 
part  at  the  left  and  below  ;  consequently  the  whole,  or  n  timos 
the  (n  -f-  l)th  term  of  the  new  series,  will  be  (m  -|-  1)  ti^^^ies 
the  sum  of  n  terms  of  the  same  series. 

We  have  already  examined  all  the  series  as  far  as  the  iburth 
order,  and  have  found  the  above  hypothesis^^  true  so  far .  Let 
us  suppose  the  series  1,  a,  &,  &c.  to  be  a  series  of  the  fourth 
order,  in  which  we  have  found  that  the  sum  of  n  term*  may  be 
obtained  by  multiplying  the  {n  +  l)th  term  by  w,  and  dividing 
the  product  by  4 ;  in  this  case  m  is  equal  to  4.  The  series 
formed  from  this  will  be  a  series  of  the  5th  order,  and  m  -\-  I 
=  4  -j-  1  =  5-  Therefore  by  the  above  demonstration  it  ap- 
pears that  the  sum  of  n  terms  of  a  series  of  the  5th  order  may 
be  obtained  by  multiplying  the  {n  -|-  l)th  term  by  «,  and 
dividing  the  product  by  5. 

If  now  the  series,  1,  a,  6,  ^Slc,  be  considered  a  series  of  the 
5th  order,  m  =  5  and  w  -|-  1  ==  6.  Hence  the  same  princi- 
ple extends  to  the  6th  order. 

If  then  we  continue  to  make  1,  a,  6,  &-c.,  represent  one  se- 
ries after  another  in  this  way,  we  shall  see  that  the  principle 
will  extend  to  any  order  whatever  of  this  kind  of  series. 

We  have  then  this  general  rule  ; 

To  find  the  sum  of  n  terms  of  a  series  of  the  order  denoted 
by  r,  derived  from  the  series  1,  1,  1,  &c.,  multiply  the  (n  -\-  \)th 
term  of  the  series  by  n  and  divide  the  product  by  r. 

Also,  the  nth  term  of  the  series  of  the  order  r,  is  equal  to  the 
sum  of  n  terms  of  the  series  of  the  order  r  —  1 . 


220  Mgebra.  XLIII. 

When  the  series  is  of  the  first  order,  the  sum  of  n  terms  is 

n.l       n 
—  or  - 

1         1 

The  sum  of  (n  -f-  1 )  terms  of  this  series  is  -^ —     This  is 

the  {n  -\-  l)th  term  of  the  series  of  the  second  order.  This 
multiphed  by  n  and  divided  by  2  gives  the  sum  of  n  terms  of 
the  series  of  the  second  order : 

n{n  -\-  \) 

1  X  2       *  * 

The  sum  of  (n  +  1)  terms  of  the  same  series  is 

{n-\-l){n^2)  1     '.     _ 

1X2 

This  is  the  (n -{-  l)th  term  of  the  series  of  the  third  order. 
This  multiplied  by  n  and  divided  by  3  gives  the  sum  of  n  terms 
of  this  series; 

71  (n  +  1)  (71 4-2) 
1X2X3 

The  sum  of  (n  +  1)  terms  of  the  last  series  is 

(n  +  l)(7i  +  2)(n+3) 
1X2X3 

This  is  the  {n  -f  l)th  term  of  the  series  of  the  fourth  order. 
This  multiplied  by  n  and  divided  by  4  gives  the  sum  of  n  terms 
of  the  series  of  the  fourth  order  : 

n{n-\-l)  (yi  +  2)  (n  +  3) 
1X2X3X4' 

Hence  for  the  series  of  the  order  r  we  have  this  formula  : 

71  (/I  +  1)  (yt  +  2)  (yi  +  3)  .  .  . .  (?i  +  r  ~  1) 
1X2X3X4X r 

We  have  examined  only  the  series  formed  from  the  series  1 , 
1,  1,  1,  &c.,  which  are  sufficient  for  our  present  purpose.  The 
principle  may  be  generalized  so  as  to  find  the  sum  of  any  series 


XLIV.  Binomial  Theorem.  221 

of  the  kind,  whatever  be  the  original  series,  and  whatever  be 
the  first  terms  of  those  formed  from  it. 

XLIV.     Binomial  Theorem. 

Before  reading  this  article,  it  is  recommended  to  the  learner 
to  review  article  XLI. 

Let  it  now  be  required  to  find  the  7th  power  of  a  -{-  a?.  The 
letters  without  the  coefficients  stand  thus ; 

a',  a*  X,  a^  x^y  a^  x^,  a^  x*,  a^  x%  a  x^,  a?'. 

The  coefficient  of  the  first  term  we  observed  Art.  XLI,  is 
always  I.  That  of  the  second  term  is  7,  the  exponent  of  the 
power,  or  the  7th  term  of  the  series  1 ,  2,  3,  &c. 

The  coefficient  of  the  third  term  is  the  sixth  term  of  the 
series  of  the  third  order  1,  3,  6, 10,  &c.  which  is  the  sum  of  six 
terms  of  the  series  1,  2,  3,  &c.  This  sum  is  found  by  multi- 
plying the  7th  term  of  the  series  by  6  and  dividing  the  product 
by  2.     But  the  7th  term  is  7,  the  coefliicient  last  found. 

121^=21. 


The  coefficient  is  21. 

The  coeflricient  of  the  fourth  term  is  the  5th  term  of  the 
series  1,  4,  10,  &c.,  or  it  is  the  sum  of  five  terms  of  the  preced- 
ing series.  The  sum  of  five  terms  of  the  series  1,  3,  6,  &c.,  is 
found  by  multiplying  the  6th  term  by  5  and  dividing  the  pro- 
duct by  3.     The  6th  term  is  the  coefficient  last  found,  viz.  21. 

5  X  21 


=  35. 


The  coefficient  is  35. 


The  coefficient  of  the  fifth  term  is  the  fourth  term  of  the 
series  of  the  fifth  order  1,  5,  15,  &c.,  or  it  is  the  sum  of  4  terms 
of  the  preceding  series.  The  sum  of  4  terms  of  the  series  1, 
4,  10,  &c.  is  found  by  multiplying  the  fifth  term  of  the  series 
by  4  and  dividing  the  product  by  4.  The  fifth  term  is  the  co- 
efficient last  found,  viz.  35. 
19* 


222  Algebra,  "XLIV. 


4 


The  coefficient  is  35. 


The  coefficient  of  the  6th  term  is  the  3d  term  of  the  series 
of  the  sixth  order,  which  is  the  sum  of  3  terms  of  the  series  of 
the  5th  order.  The  sum  of  3  terms  of  this  series  is  found  by 
multiplying  the  4th  term  by  3  and  dividing  the  product  by  5. 
The  4th  term  is  the  coefficient  last  found,  viz.  35 

5 

The  coefficient  is  21. 

The  coefficient  of  the  7th  term  is  the  2d  term  of  the  series 
of  the  7th  order,  which  is  the  sum  of  two  terms  of  the  series  of 
the  6th  order.  The  3d  term  of  this  series  is  the  coefficient  last 
found,  viz.  21. 

2  X  21       . 


The  coefficient  is  7. 

The  coefficient  of  the  last  term  is  1,  though  it  may  be  found 
by  the  rule 

1  X  7  ^  J 

7 

Hence  the  7th  power  of  «  -|-  jc  is 

a'  J\-l  a^  X  ^21  a'x"  -\-^b  a'x''  ^^ba^  x'  -\-2\  a^  x''  -{-1  ax^-^x' 

Examining  the  formation  of  the  above  coefficients,  we  ob- 
serve, that  each  coefficient  was  found  by  multiplying  the  coef- 
ficient of  the  preceding  term  by  the  exponent  of  the  leading 
quantity  a  in  that  term,  and  dividing  the  product  by  the  num- 
ber which  marks  the  place  of  that  term.  Thus  the  coefficient 
of  the  third  term  was  found  by  multiplying  7,  the  coefficient  of 
the  second  term,  by  6,  the  exponent  of  a  in  the  second  term, 
and  dividing  the  product  by  2,  the  number  which  marks  the 
place  of  the  second  term.  This  will  be  true  for  all  cases,  be- 
cause that  exponent  must  necessarily  show  the  number  of  terms 
of  which  the  sum  is  to  be  found;  the  coefficient  will  always  be 


XLIV.  •      Binomial  Theorem,  223 

the  term  to  be  multiplied,  because  the  number  of  terms  al- 
ways diminishes  by  1  for  the  successive  coefficients,  and  the 
place  of  the  term  always  marks  the  order  of  the  series  of  which 
the  sum  is  to  be  found. 

Hence  is  obtained  the  following  general  rule. 

Knovnng  the  coefficient  of  any  term  in  the  power,  the  coefficient  of 
the  succeeding  term  is  found  by  multiplying  the  coefficient  of  the 
known  term  by  the  exponent  of  the  leading  quantity  in  that  term, 
and  dividing  the  product  by  the  number  which  marks  the  place  of 
that  term  from  the  first. 

The  coefficient  of  the  first  term,  being  always  1,  is  always 
known.  Therefore,  beginning  with  this,  all  the  others  may  be 
found  by  the  rule. 

It  may  be  farther  observed,  that  the  coefficients  of  the  last 
half  of  the  terms,  are  the  same  as  those  of  the  first  half  in  an 
inverted  order.  This  is  evident  by  looking  at  the  coefficients, 
page  275,  and  observing  that  the  series  are  the  same,  whether 
taken  obliquely  to  the  left  or  to  the  right. 

It  is  also  evident  from  this,  that  a-^  xis  the  same  as  x  -\-  a, 
and  that,  taken  fi-om  right  to  left,  x  is  the  leading  quantity  in 
the  same  manner  as  a  is  the  leading  quantity  from  left  to 
right. 

Hence  it  is  sufficient  to  find  coefficients  of  one  half  of  the 
terms  when  the  number  of  terms  is  even,  and  of  one  more  than 
half  when  the  number  is  odd.  The  same  coefficients  may  then 
be  written  before  the  corresponding  terms  counted  from  the 
right. 

In  the  above  example  of  the  6th  power,  the  coefficients  of 
the  first  four  terms  being  found,  we  may  begin  on  the  right 
and  put  6  before  the  second,  and  15  before  the  third,  and  then 
the  power  is  complete* 

Examples.  ' 

1.  What  is  the  7th  power  ofa-\-x? 

Jins.  a'  -f  7  a«  a?  -1-21  a'  a?"  +  35  a^ip'  -f  35  a' a;*  +  21  a' a?' 

>  ^  +lax^-\-x\ 


224  V         Jllgehra.  XLIV. 

2.  What  is  the  10th  power  of  a  +  a:  ? 

Am.  a'*  +  10a»a?  +  45fl«a^'+  120a'cc' +  210a«a?* -f  . . . 
252  a^  x^  +  210  a*  a:«  +  120  a^  x'  +  45  a''  a?«  +  10  a  a?'  -f  ^'°- 

3.  What  is  the  9th  power  of  a  +  5  ? 

4.  What  is  the  13th  power  of  wi  +  ^  •'' 

5.  What  is  the  2d  power  of  2  a  c  +  c?  f 
Make  2  ac=  b. 

The  2d  power  of  b -{-d  is  h^  -j- 2b d^d\ 

Putting  2  a  c  the  value  of  b  into  this,  instead  of  b,  observing 
that  b^  =z4a^  c^  and  it  becomes 

4a'c'  +  4acJ  +  £?*. 

6.  What  is  the  3d  power  of  3  c'  +  2  5  <?  ? 
Make  a:=Sc^  and  x  =  2bd. 

The  3d  power  of  a  -\-  x  is  a^  -^  3  a^  x  -{-  3  a  x^  -{-  x^. 
Put  into  this  the  values  of  a  and  x  and  it  becomes 
27c'-{-54c'bd  +  36  c'  b'd'-{-8  b^  ^, 
which  is  the  3d  power  of3c^-\-2bd. 

7.  What  is  the  3d  power  of  a  —  6  ? 

Make  x  =  —  b,  then  having  found  the  3d  power  of  a  -|-  a? 
put  —  6  in  the  place  of  x  and  it  becomes 

a'^Sa'b  +  3ab'^b\ 

which  is  the  3d  power  of  cf  —  b. 

In  fact  it  is  evident  that  the  powers  of « —  b  will  be  the 
same  as  the  powers  of  a  +  6,  with  the  exception  of  the  signs. 
It  is  also  evident  that  every  term  which  contains  an  odd  power 
of  the  term  affected  with  the  sign  —  must  have  the  sign  — , 
and  every  term  which  contains  an  even  power  of  the  same 
quantity  must  have  the  sign  +. 

8.  What  is  the  7th  power  of  m  —  w  ? 

9.  What  is  the  4th  power  of  2  a  —  bc^? 

10.  What  is  the  5th  power  of  a^  c  —  2  c*? 


XLIV.  Binomial  Theorem.        ,  225 

11.  What  is  the  3d  power  of  a  -f  5  -f-  c  ? 

Make  m  =:  6  -|-  c.     Then  a-\-m-=^a-\-h'\-c. 

The  3d  power  of  «  -f-  ^  is  a^  -|-  3  a'  m  -|-  3  a  m*  -|-  m'. 

But  wi  =  6  -j-  c,  m'  =  6*  -|-  2  6  c  -j-  c*,  and 

m'=5'  +  3  6^c  +  3  6c'-|-c^ 

Substituting  these  values  of  w,  the  third  power  of  a  -|-  ^  +  c 
will  be 

a='+3«'6+3aV+3a2>'+6«6c-|-3ac'+J'+36'c4-36c''-|-c^ 

%  12.  What  is  the  3d  power  of  «  —  h  -{- c^. 

Make  a  —  h  z-m,  raise  m  -J-  c  to  the  3d  power,  and  then  sub- 
stitute the  value  of  m. 

Am.  a'  — 3a'5  +  3a'c  +  3a6'-— 6a6c  — 3ac'  — 6\... 

which  is  the  same  as  the  last,  except  that  the  terms  which  con- 
tain the  odd  powers  of  h  have  the  sign  — . 

Hence  it  is  evident  that  the  powers  of  any  compound  quan- 
tity whatever,  may  be  found  by  the  binomial  theorem,  if  the 
quantity  be  first  changed  to  a  binomial  with  two  simple  terms, 
one  letter  being  made  equal  to  several,  that  binomial  raised  to 
the  power  required,  and  then  the  proper  letters  restored  in  their 
places. 

13.  What  is  the  2d  power  ofa-f^>-|-c  —  c?? 
Ms.a^'\-2ah-\-¥  +  2ac-{-2hc  —  2ad'-^2hd+e 

—  2crf  +  (^^ 

14.  What  is  the  3d  power  of  2  a  —  ^  +  c*  ? 

15.  What  is  the  7th  power  of  3  a'  —  2  a'  rf  ? 

16.  What  is  the  4th  power  of  7  6'  +  2  c  —  (?'  ? 

17.  What  is  the  13th  power  of  a*  —  2  i'  ? 

18.  What  is  the  5th  power  of  a'  ~  c  —  2  <^? 

19.  What  is  the  3d  power  ofa  —  2<?4-c*(/? 

20.  What  is  the  3d  power  of  a  —  6  —  2  c''  —  <f? 

21.  What  is  the  5th  power  of  7  a'  ¥  —  10  a'  c'? 


3^  Algebra.  XLV. 


XLV.  The  rule  for  finding  the  coefficients  of  the  powers  of 
binomials  may  be  derived  and  expressed  more  generally  as 
follows : 

It  is  required  to  find  the  coefficients  of  the  nth  power  of 

a  -{-  X. 

It  has  already  been  observed,  Art.  XLL,  that  the  coefficient 
of  the  second  term  of  the  nth  power  is  the  ?ith  term  of  the  se- 
ries of  the  second  order,  1,  2,  3,  &c.,  or,  the  sum  of  n  terms 
of  the  series  1,  1,  1,  &c. ;  that  the  coefficient  of  the  third 
term  is  the  sum  of  (w  —  1)  terms  of  the  series  of  the  second 
order ;  that  the  coefficient  of  the  fourth  term  is  the  sum  of 
(n  —  2)  terms  of  the  series  of  the  third  order,  &c.  So  that 
the  coefficient  of  each  term  is  the  sum  of  a  number  of  terms 
of  the  series  of  the  order  less  by  one,  than  is  expressed  by  the 
place  of  the  term  ;  and  the  number  of  terms  to  be  used  is  less 
by  one  for  each  succeeding  series. 

By  Art.  XLIL  the  sum  of  n  terms  of  the  series  1,  1,1,  is 
I  The  sum  of  [n—  1), terms  of  the  series  of  the  second 
order  is 

71  (n  — .  1) 
1  X  2       * 

The  sum  of  {n  —  2)  terms  of  the  series  of  the  third  order  is 

n{n—l)  (n-— 2) 
1X2X3 

Hence  (a  +  a:)n  =  ««  _|-  !^  a""^  x  -f  ^(^  —  ^)  a  «-*  x' 
^  ^  ^1  1X2 

^        n(n-l)  (n-2)   ^  „_3  ^,3    ,    ^^ 
1X2X3 

It  may  be  observed  that  n  is  the  exponent  of  a  in  the  first 

term,  and  that  n  or  its  equal  -  forms  the  coefficient  of  the  se- 
cond term.  1 


XLV.  Binomial  Theorem.  221 

The  coefficient  of  the  third  term  is  ^  multiplied  by  !!!II1-,  or 

multiplied  by  (n  —  1)  and  divided  by  2.  But  [n  —  1)  is  the 
exponent  of  a  in  the  2d  term,  and  2  marks  the  place  of  the 
second  term  from  the  left.  Therefore  the  coefficient  of  the 
third  term  is  found  by  multiplying  the  coefficient  of  the  second 
term  by  the  exponent  of  a  in  that  term,  and  dividing  the  pro- 
duct by  the  number  which  marks  the  place  of  that  term  from 
the  left. 

By  examining  the  other  terms,  the  following  general  rule 
will  be  found  true. 

Multiply  the  coefficient  of  any  term  by  the  exponent  of  the  lead- 
ing quantity  in  that  term,  and  divide  the  product  by  the  number  that 
rtmrks  the  place  of  that  term  from  the  left,  and  you  mil  obtain  the 
coefficient  of  the  next  succeeding  term.  Then  diminish  the  exponent 
of  the  leading  quantity  by  1  and  increase  that  of  the  other  by  1  and 
the  term  is  complete. 

By  this  rule  only  the  requisite  number  of  terms  can  be  ob- 
tained. For  x"^,  which  is  properly  the  last  term  of  (a  +  a?)", 
is  the  same  as  a"  x^.  If  we  attempt  by  the  rule  to  obtain  ano- 
ther term  from  this,  it  becomes  0  X  a~^  x  "+'  which  is  equal  to 
zero. 

It  has  been  remarked  above,  that  the  coefficients  of  the  last 
half  of  the  terms  of  any  power,  are  the  same  as  those  of  the 
first  reversed.     This  may  be  seen  from  the  general  expression : 

If  w  =  7,  then  _  =  _ ; =  -  ; =  _ ; 

1        l'       2  2'       3  3' 

n  —  ^__A__^^    n-—4       3  .  w  — -  5  _  2 , 


4  4  '         5 


1 


>  — ^ z  » 


7  7 


This  furnishes  the  following  fractions,  viz. 


7        6        5        4        3        2         1 
1>     2  J     ¥>     4>     5>     e"?     T* 


226  Algebra,  XLVI. 

The  first  of  these  is  the  coefficient  of  the  second  term  ;  the 
coefficient  of  the  second  multipfied  by  f  forms  the  coefficient 
of  the  third  term,  <fec. 


21.     21  X  4  —  35. 


Now  35  multiplied  by  |  =  1  will  not  be  altered  ;  hence  two 
successive  coefficients  will  be  alike.  21  multiplied  by  f  pro- 
duced 35  ;  so  35  multiplied  by  f  must  reproduce  21.  In  this 
way  all  the  terms  will  be  reproduced ;  for  the  last  half  of  the 
fractions  are  the  first  half  inverted. 

This  demonstration  might  be  made  more  general,  but  it  is 
not  necessary. 


XLVI.     Progression  by  Difference,  or  Arithmetical  Progression. 

A  series  of  numbers  increasing  or  decreasing  by  a  constant 
difference,  is  called  a  progression  by  difference,  and  sometimes 
sn  arithmetical  progression. 

The  first  of  the  two  following  series  is  an  example  of  an  in- 
creasing, and  the  second  of  a  decreasing,  progression  by  dif- 
ference. 

5,       8,     11,     14,     17,     20,     23 

50,     45,     40,     35,     30,     25,     20 

It  is  easy  to  find  any  term  in  the  series  without  calculating 
the  intermediate  terms,  if  we  know  the  first  term,  the  common 
difference,  and  the  number  of  that  term  in  the  series  reckoned 
from  the  first. 

Let  a  be  the  first  term,  r  the  common  difference,  and  n  the 
number  of  terms.     The  series  is 

«,a  4-r,  a  -|-2r,  a  -f  3r a-\-  {n  —  2)  r,  a -{-  {n —  l)r. 

The  points are  used  to  show  that  some  terms  are  left 

out  of  the  expression,  as  it  is  impossible  to  express  the  whole 
until  a  particular  value  is  given  to  n. 

Let  I  be  the  term  required,  then 

/=:a  +  (n — l)r. 


XLVI.  Progression  by  Difference,  229 

Hence,  any  terai  may  be  found  by  adding  the  product  of  the 
common  difference  by  the  number  of  terms  less  one,  to  the  first 
term. 

Example. 

What  is  the  10th  term  of  the  series  3,  5,  7,  9,  &c. 
In  this  a  =  3,  r  =  2,  and  n  —  1  =9. 

/rr:3+9'X2  =  21.  ' 

In  a  decreasing  series,  r  is  negative. 
Example. 

What  is  the  13th  term  of  the  series  48,  45,  42,  &c.  f 

«  =  48,  r  ::=:  —  3,  and  n  —  1  =  12. 

Z  =  48  + (12  X —3)=  48  — 36  =  12. 

Let  a,  h,  c,  be  any  three  successive  terms  in  a  progression  by 
difference. 

By  the  definition,  ^ 

h — a  =.c  —  b 

2b  =a-j-c 

b  =  l±I. 

2 

That  is,  if  three  successive  terms  in  a  progression  by  differ- 
ence be  taken,  the  sum  of  the  extremes  is  equal  to  twice  the 
mean. 

Example. 

Lqt  the  three  terms  be  3,  5,  and  7. 

2x5  =  7+3=  10. 

Example  2d.  Let  7  and  17  be  the  first  and  last  term,  what 
is  the  mean  ? 

20 


230  Algebra,  XLVL 

.:  =  L±1!=:12. 
2 

Let  a,  Z>,  c,  <?,  be  four  successive  terms  of  a  progression  by 
difference* 

h —  a  =zd  —  c 

b-\-c=za  +  d. 

That  is,  the  sum  of  the  two  extremes  is  equal  to  the  sum  of 
the  two  means. 

Example. 

Let  5,  9,  13, 17,  be  four  successive  terms. 
9+13=17  +  5=22. 

Let  a,b,c,  d,e h,  i,  k,  I,  be  any  number  of  terms  in  a 

progression  by  differences  ;  by  the  definition  we  have 

b  —  a  =  I  —  k 

c  —  b  zzzk  —  i 

d  —  c=:i  —  h,  &,c, 

which  by  transposition  give 

a+  I  =b+k, 

6  +  A;  =  c  +  1. 

c  +  i  z=z  d  -\-  h,  &c. 

That  is,  if  the  first  and  last  be  added  together,  the  second 
and  the  last  but  one,  the  third  and  the  last  but  two,  the  sums 
will  all  be  equal. 

Example, 
Let  3,  5,  7,  9,  11,  13,  be  such  a  series. 


XLVI.  Progression  by  Difference,  2S1 

3,       5,       7,       9,     11,     13, 
13,     11,       9,       7,       5,       3, 


16,     i6,     16,     16,     16,     16. 


It  will  now  be  easy  to  find  the  sum  of  all  the  terms  in  any 
progression  by  difference,  and  that  even  when  but  part  of  the 
terms  are  known. 

Let  S  represent  the  sum  of  the  series,  then  we  have 

S=za+b-\-c-i-d-^.,..h-\-i  +  k-\-l 

Also  S  =  l  +  k  +  i+h  +  ....d-{-c  +  b  +  a. 

Adding  these  term  to  term  as  they  stand, 

2S=(a+0  +  (*  +  fc)  +  (c  +  i)  +  («i  +  A)  +  -...(d  +  A)  +  (c  +  i)  +  (ft+fc)  +  (a  +  0, 

But  it  has  just  been  shown  that 

a'\-l  =  b'\-k-zc-]-i,  &c. 

That  is,  all  the  terms  are  now  equal,  and  one  of  them  be- 
ing multiplied  by  the  whole  number  of  terms  will  give  the 
whole  sum  :  thus 

2S  =  n{a-^l) 

2 

Hence,  the  sum  of  a  series  of  numbers  in  progression  by  differ- 
ence is  one  half  of  the  product  of  the  number  of  terms  by  the  sum  of 
the  first  and  last  terms. 


Example, 

How  many  strokes  does  the  hammer  of  a  clock  strike  in  12 
hours  ? 

a  =  1,     /  =  12,     and  n  =  12. 

S  =  i?il-±i^  =  78.     ^n*.  78  strokes. 

2 

In  the  formula  Z  =  «  -f-  (n  —  1)  r;  sub..titute  d  instead  of 
r  to  represent  the  difference ;  thus 

I  z=:  a  -{-  (n  —  l)eif. 


232  Algebra.  XLVI. 

This  formula  and  the  following 

f^  _n{a  +  l) 

2      ' 

contain  five  different  things,  viz.  «,  /,  w,  </,  and  S ;  any  three 
of  which  being  given,  the  other  two  may  be  found,  by  combin- 
ing the  two  equations.  I  shall  leave  the  learner  to  trace  these 
himself  as  occasion  may  require. 

Examples  in  Progression  by  Difference. 

1.  How  many  strokes  do  the  clocks  of  Venice,  which  go  on 
to  24  o'clock,  strike  in  a  day  ^ 

2.  Suppose  100  stones  to  be  placed  in  a  straight  line  3  yards 
asunder  ;  how  far  would  a  person  travel  who  should  set  a  bas- 
ket 3  yards  from  the  first,  and  then  go  and  pick  them  up  one 
by  one,  and  put  them  into  the  basket  ? 

3.  After  A,  who  travelled  at  the  rate  of  4  miles  an  hour,  had 
been  set  out  2f  hours,  B  set  out  to  overtake  him,  and  in 
order  thereto  went  four  miles  and  a  half  the  first  hour,  four  and 
three  fourths  the  second,  five  the  third,  and  so  on,  increasing 
his  rate  one  fourth  of  a  mile  e^ph  hour.  In  how  many  hours 
will  he  overtake  A  ? 

The  above  example  is  solved  by  using  both  the  above  for- 
mulas. The  known  quantities  are  the  first  term,  the  difference, 
and  the  sum  of  all  the  terms.  The  unknown  are  the  last  term, 
and  the  number  of  terms.  It  involves  an  equation  of  the  se- 
cond degree.  It  is  most  convenient  to  use  x,  y,  &c.  for  the 
unknown  quantities. 

4.  A  and  B  set  out  from  London  to  go  round  the  world, 
(24990  miles,)  one  going  East  and  the  other  West.  A  goes 
one  mile  the  first  day,  two  the  second,  three  the  third,  and  so 
on,  increasing  his  rate  one  mile  per  day.  B  goes  20  miles  a 
day.  In  how  many  days  will  they  meet,  and  how  many  miles 
will  each  travel  ? 

5.  A  traveller  sets  out  for  a  certain  place,  and  travels  1  mile 
the  first  day,  2  the  second,  and  so  on.  In  5  days  afterwards 
another  sets  out,  and  travels  12  miles  a  day.  How  long  and 
how  far  must  he  travel  to  overtake  the  first .'' 


XL  VII.  Geometiical  Progression.  233 

6.  A  and  B  165  miles  distant  from  each  other  set  out  with  a 
design  to  meet ;  A  travels  1  mile  the  first  day,  2  the  second, 
3  the  third,  and  so  on.  B  travels  20  miles  the  first  day;  18  the 
second,  16  the  third,  and  so  on.     How  soon  will  they  meet  ? 

Ans.  They  will  be  together  on  the  10th  day,  and  continuing 
that  rate  of  travelling,  they  may  be  together  again  on  the  33d 
day.     Let  the  learner  explain  how  this  can  take  place. 

7.  A  person  makes  a  mixture  of  51  gallons,  consisting  of 
brandy,  rum,  and  water  ;  the  quantities  of  which  are  in  arith- 
metical progression.  The  number  of  gallons  of  brandy  and 
rum  together,  is  to  the  number  of  gallons  of  rum  and  water 
together  as  8  to  9.     Required  the  quantities  of  each. 

Let  X  =z  the  number  of  gallons  of  rum 

and  y  =.  the  common  difference. 

Then  x  —  y,  x,  and  x  -{-  y  will  express  the  three  quantities. 

8.  A  number  consisting  of  three  digits  which  are  in  arith- 
metical progression,  being  divided  by  the  sum  of  its  digits,  gives 
a  quotient  48  ;  and  if  198  be  subtracted  from  the  number,  the 
digits  will  be  inverted.     Required  the  number. 

9.  A  person  employed  3  workmen,  whose  daily  wages  were 
in  arithmetical  progression.  The  number  of  days  they  worked 
was  equal  to  the  number  of  shillings  that  the  second  received 
per  day.  The  whole  amount  of  their  wages  was  7  guineas,  and 
the  best  workman  received  28  shillings  more  than  the  worst. 
What  were  their  daily  wages  ? 

Progression  by  difference  is  only  a  particular  case  of  the 
series  by  difference,  explained  Arts.  XL.  and  XLL  All  the 
principles  and  rules  of  it  may  be  derived  from  the  formulas 
obtained  there.  It  would  be  a  gTOd  exercise  for  the  learner 
to  deduce  these  rules  from  those  formulas. 

^-  '^     '^    '" 

XLVII.     Progression  by  Quotient,  or  Geometrical  Progression. 


Progression  by  quotient  is  a  series  of  numbers  such,  that  if 
any  term  be  divided  by  the  one  which  precedes  it,  the  quotient 
is  the  same  in  whatever  part  the  two  terms  be  taken.     If  the 
20* 


234  dlgcbra,  XLVIL 

series  is  increasing,  the  quotient  will  be  greater  than  unity,  if 
decreasing,  the  quotient  will  be  less  than  unity. 

The  following  series  are  examples  of  this  kind  of  progres- 
sion. 

3,  6,     12,     24,     48....&C. 
72.     24         8         1         8        8    • 

In  the  first  the  quotient  (or  ratio,  as  it  is  generally  called,)  is 
2,  in  the  second  it  is  ^. 

Let  a,b,  c,d, ,. . .  k,  I,  be  a  series  of  this  kind,  and  let  g  re- 
present the  quotient. 

Then  we  have  by  the  definition, 

_6_c_</_e__         / 
a       h       c        d  '   k 

From  these  equations  we  derive 

h  =:  aq,  c  :=:  b  q,  d  =:  c  q,  e  =z  d  q  ,..'..  I  =z  k  q. 

Putting  successively  the  value  of  b  into  that  of  c,  and  that  of 
f  into  that  of  (?,  &,c.,  they  become 

b  =:z  aq,  c:=^  aq^,  d  =^  aq^,  c  :=  a  q*,  .  .  .  .  I  =  a  q  *""', 

designating  by  n,  the  rank  of  the  term  /,  or  the  number  of  terms 
in  the  proposed  progression. 

Any  term  whatever  in  the  series  may  be  found  without  find- 
ing the  intermediate  terms,  by  the  formula 

lmmaq''-\ 

4.  Example.         *^ 

What  is  the  7th  term  of  the  series  3,  6,  12,  &c.  ^ 
Here  «  =r  3,  9  =  2,  and  n  —  1=6. 

/r=3  X  2**=  192.  Am.  192. 


XL VII.  Geometrical  Progression.  235 

We  may  also  find  the  sum  of  any  number  of  terms  of  the 
progression 

a,  6,  c,  d,  (fcc. 

If  we  add  the  equations 

b  =  aq,  c=.bq,  d  =i  c q,  e=:  dq I  =.kq, 

we  obtain 

b'^c-{-d-\-e-^ l={a-\-b-\-c-{'d-\-e-{- k)  q. 

Observe  that  the  first  member  is  the  sum  of  all  the  terms  of 
the  progression  except  the  first,  a,  and  the  part  of  the  second 
member  enclosed  in  the  parenthesis,  is  the  sum  of  all  the  terms 
except  the  last,  I ;  and  this,  multiplied  by  9,  is  equal  to  the 
first  member. 

Now  putting  S  for  the  sum  of  all  the  terms,  we  have 

J-j-c  +  df-fe-f l^zS'-a 

«  +  ^-f-c  +  <?H-e-|- k  =  S^L 

Hence  we  conclude  that 

S^a  =  (S-^l)q, 


which  gives 


Example,        ^ 

What  is  the  sum  of  seven  terms  of  the  series 
5,  15,  45,  &c. 
l=zdxS'=:  3645 

S  =  ^X^^^^-^  =  5465. 
3  —  1 

The  two  equations 

I  =  aq'^-',  B.nd  S  :=zlLzJ!:, 
9—1 


236  Algebra,  XLVII. 

contain  all  the  relations  of  the  five  quantities  a,  I,  q,  n,  and  S ; 
any  three  of  which  being  given,  the  other  two  may  be  found. 
It  would  however  be  difficult  to  find  n,  without  the  aid  of  lo- 
garithms, which  will  be  explained  hereafter.  Indeed  loga- 
rithms will  greatly  facilitate  the  calculations  in  most  cases  of 
geometrical  progression.  Therefore  we  shall  give  but  few  ex- 
amples, until  we  have  explained  them. 

If  we  substitute  a  q^~^  in  place  of  Z,  in  the  expression  of  S,  it 
becomes 

.«;_«(g°-l) 

When  q  is  greater  than  unity,  the  quantity  q^  will  become 
greater  as  n  is  made  greater,  and  jS  may  be  made  to  exceed 
any  quantity  we  please,  by  giving  n  a  suitable  value  ;  that  is, 
by  taking  a  sufficient  number  of  terms.  But  if  g  is  a  fraction 
less  than  unity,  the  greater  the  quantity  w,  the  smaller  will  be 

the  quantity  q^.     Suppose  q  =:  — ,  m  being  a  number  greater 

m 

than  unity,  then 

wiP- 

Substituting  —  in  place  of  (f  in  the  expression  of  »S,  and  it 
becomes 


.(?-) 


i-i 


m 


Changing  the  signs  of  the  numerator  and  denominator,  and 
multiplying  both  by  m. 


\\  —  —  I        am  —  — 

\        irCf  —  n 


am\\  —  —  I       am  — am 

s= - 


m  —  1  m  —  1  m  —  1* 

It  is  evident  that  the  larger  n  is  or  the  more  terms  we  take 


XLVII.  Geometrical  Progression*  '  237 

in  the  progression,  the  smaller  will  be  the  quantity  — ^,  and 


7W» 


am 


consequently  the  nearer  the  value  of  5  will  approach 

m  —  1' 

from  which  it  differs  only  by  the  quantity 


(m — l)w"~^ 
But  it  can  never,  strictly  speaking,  be  equal  to  it,  for  the 
quantity  - will  always  have  some  value,  however 

large  n  may  be  ;  yet  no  quantity  can  be  assumed,  but  this  ex- 
pression may  be  rendered  smaller  than  it. 

The  quantity     ^^      is  therefore  the  limit  which  the  sum 
m  —  1 

of  a  decreasing  progression  can  never  surpass,  but  to  which 

the  value  continually  approximates,  as  we  take  more  terms  in 

the  series. 

In  the  progression 
^         •  1,  h  h  h  tV>  &c- 

m 
_    1 X2  1  _^  1 


Hence  S  =  JLiLf  — i =  2-~ 


2  —  1       (2  —  1)  X  2"-'  1  X  2«-* 

In  this  example  the  more  terms  we  take,  the  nearer  the  sum 
of  the  series  will  approach  to  2,  but  it  can  never  be  strictly 
equal  to  it.     Now  if  we  consider  the  number  of  terms  infinite, 

the  quantity will  be  so  small  that  it  may  be  omitted 

-^  1  X  2«-'  •" 

without  any  sensible  error,  and  the  sum  of  the  series  may  be 
said  to  be  equal  to  2. 

By  taking  more  and  more  terms  we  approach  2  thus, 


238  Algebra.  XLVII. 

1  =  2  —  1 

1+i  =2-i 

1+i  +  i  =2— i 

l+*+4+i  =2-1 

l+J  +  i  +  l  +  TV=  2— ^V,&c. 

Examples. 

What  is  the  sum  of  the  series  1,  J,  ^,  J^f,  ifcc.  continued  to 
an  infinite  number  of  terms  ? 


wi       3 


lX3^3^jj 


3—1 


2.  What  is  the  sum  of  the  series,  5,  f,  |,  ^\,  &c.  continued 
to  an  infinite  number  of  terms  f 

3.  What  is  the  sum  of  the  following  series  continued  to  in- 
finity ? 

36,  7,  1,  /^,  &c. 

4.  What  is  the  sum  of  the  following  series  continued  to  in- 
finity ? 

208,26,34,  if,  <Slc. 

5.  What  is  the  sum  of  the  following  series  continued  to  in- 
finity ? 

38,  4f,  li|.  If,  &-C. 

6.  What  is  the  10th  term  of  the  series 

5,  15,45,  &c.? 

7.  What  is  the  sum  of  8  terms  of  the  series 

35,  175,  875,  &c.  ? 

When  three  numbers  are  in  geometrical  prc^ression,  the 
middle  term  is  called  a  mean  proportional  between  the  other 
two. 


XLVUI.  Logarithms.  239 

Let  three  numbers,  a,  b,  c,  be  in  geometrical  progression,  so 
that 


a         b 

We  have 

6*  =  ac 

and 

hz={ac)^. 

8    Find  a  mean 

proportional  between  4  and  9. 

4  _  a? 

1^"  "9 

x'=.m 

^   =:     6. 

9.  Find  a  mean 

proportional  between  7  and  10. 

10.  Find  a  mean  proportional  between  2  and  3, 

Am.  6. 


XLVllI.     Logarithms, 

We  have  seen,  Art.  XXXVIII,  with  what  facility  multiplica- 
tion, division,  the  raising  of  powers,  and  the  extraction  of  roots 
may  be  performed  on  literal  quantities  consisting  of  the  same 
letter,  by  operating  on  the  exponents.  We  propose  now  to 
apply  the  same  principle,  though  in  a  way  a  little  different,  to 
numbers. 

Multiplication,  we  observed,  is  performed  by  adding  the  ex- 
ponents, and  division  by  subtracting  the  exponent  of  the  divisor 
from  that  of  the  dividend. 


Thus  d'  X  a'  is  «'+'  =  a\     And  —  is  a'-'  =  a\ 

a' 

In  the  same  manner  2'  x  2'  =  2'+'  =  2^ 

and  ~  =  2'-'  =  2\ 

2' 

Let  us  make  a  table  consisting  of  two  columns,  the  first  con- 
taining the  different  powers  of  2,  and  the  second  the  exponents 
of  those  powers. 


240 


Algebra. 


XLVIIL 


Observe  first  that  a°  =  1,  so  also  2°  =  1,  2'  =  2,  2'  =  4, 
2'  =  8,  2*  =  16,  2*^  =  32,  2*  =  64,  2'  =  128,  &c. 


TABLE. 


Powers. 

1 

2 

4 

8 
16 
32 
64 


Expon. 

Powers. 

Expon. 

0 

128 

7 

1 

256 

8 

2 

612 

9 

3 

1024 

10 

4 

2048 

11 

5 

4096 

12 

6 

8192 

13     ! 

Powers. 

16,384 

32,768 

65,536 

131,072 

262,144 

524,288 

1,048,576 


Expon. 
14 
15 
16 
17 
18 
19 
20 


Suppose  now  it  is  required  to  multiply  256  by  64.  We  find 
by  the  table  that  256  is  the  8th  power  of  2,  that  is  2^  and  that 
64  is  T.  Now  2«  X  2«  =1  2^+^  —  2'\  Returning  to  the  table 
again  and  looking  for  14  in  the  column  of  exponents,  against 
it  we  find  16384  for  the  14th  power  of  2.  Therefore  the  pro- 
duct of  256  by  64  is  16384. 


This  we  may  easily  prove. 


256 
64 


1024 
1536 


16384 
Multiply  256  by  128. 

Finding  these  numbers  in  the  table  in  the  column  of  powers, 
and  looking  in  the  other  column  for  the  exponents,  we  find 
that  256  is  the  8th  power  of  2,  and  128  the  7th  power.  Add- 
ing the  exponents  8  and  7,  we  have  15  for  the  exponent  of 
the  product.  Now  looking  for  1 5  in  the  column  of  exponents, 
we  find  against  it  in  the  column  of  powers,  32768  for  the  15th 
power  of  2,  which  is  the  product  of  256  by  128.  Let  the 
learner  prove  this  by  multiplying  256  by  128. 

Divide  8192  by  32. 

Looking  for  these  numbers  in  the  column  of  powers,  and  for 


XLVIII.  Logarithms,  241 

the  corresponding  exponents,  we  find  8192  is  the  1 3th  power 
of  2,  and  32  is  the  5th  power. 

__  zzz  2}^^^  ::=:  2' 

2* 

Looking  for  8  in  the  column  of  exponents,  and  for  its  corre- 
sponding number,  we  find  256  for  the  8th  power  of  2,  or  the 
quotient  of  8 192  by  32. 

Divide  32768  by  512. 

The  exponents  corresponding  to  these  numbers  in  the  table 
are  15  and  9.  15  —  9  =1  6.  In  the  column  of  exponents,  6 
corresponds  to  64,  which  is  the  true  quotieVit  of  32768  by  512 

What  is  the  3d  power  of  32  ? 

The  exponent  corresponding  to  32  is  5.  Now  to  find  the 
3d  power  of  a^  we  should  multiply  the  exponent  by  3,  thus 
rt  5  X  3  -_  ^15^  So  the  third  power  of  2'  is  2  ^  x  3  z=:  2'\  Against 
15  in  the  column  of  exponents  we  find  32768  for  the  15th 
power  of  2.     Therefore  the  3d  power  of  32  is  32768. 

What  is  the  2d  power  of  128  ? 

The  exponent  corresponding  to  this  number  is  7.  7x2  = 
14.  The  number  corresponding  to  the  exponent  14  is  16384, 
which  is  the  second  power  of  128. 

What  is  the  3d  root  of  4096  ? 

The  exponent  corresponding  to  this  number  is  12. 

The  3d  root  of  2''  is  2  '"^  =  2\ 

The  number  corresponding  to  the  exponent  4  is  16,  which 
is  the  3d  root  of  4096. 

What  is  the  fourth  root  of  65,536  ? 

The  exponent  corresponding  to  this  number  is  16,  which 
divided  by  4  gives  for  the  exponent  of  the  root  4,  the  number 
corresponding  to  which  is  16.     The  answer  is  16. 

21 


242  '  Algebra.  XLIX. 

Examples. 

1.  Multiply  512  by  256. 

2.  Multiply  8192  by  128. 
*  3.  Multiply  2048  by  256. 

4.  Divide  262,144  by  128. 

5.  Divide  1,048,576  by  512. 

6.  Divide  524,288  by  131,072. 

7.  What  is  the  2d  power  of  1024 .? 

8.  What  is  the  3d  power  of  64  ? 

9.  What  is  the  5th  power  of  16  ? 

10.  What  is  the  2nd  root  of  262,144  ^ 

11.  What  is  the  3d  root  of  262,144 .? 

12.  What  is  the  4th  root  of  1,048,576  ? 

13.  What  is  the  5th  root  of  1,048,576  .? 

14.  What  is  the  6th  root  of  262,144  ? 

The  operations  of  multiplication,  division,  and  the  extrac- 
tion of  roots  are  very  easy  by  means  of  this  table.  This  table 
however  contains  but  very  few  numbers.  But  an  exponent  of 
2  may  be  found  for  all  numbers  from  1  as  high  as  we  please. 
For  2'  =  2,  and  2^  =:  4.  Hence  the  exponent  of  2  answering 
to  the  number  3  will  be  between  1  and  2 ;  that  is,  1  and  a 
fraction.  So  the  exponents  answering  to  5,  6,  and  7,  will  be  2 
and  a  fraction,  &.c. 

XLIX.  A  table  may  also  be  made  of  the  powers  of  3,  or  of 
4,  or  any  other  number  except  1 ,  which  shall  have  the  same 
properties.  Exponents  might  be  found  answering  to  every 
number  from  1  upwards. 

3°  =  1,3^  =  3,  3^  =  9,  3'  =  27,  &c. 

The  column  of  powers  will  always  consist  of  the  numbers  1, 
2,  3,  &c.  but  the  column  of  exponents  will  be  different  accord- 
ing as  the  numbers  are  considered  powers  of  a  different  num- 
ber^ 


XLIX.  Logarithms.  243 

The  formula  a^  =y  will  apply  to  every  table  of  this  kind. 

If  any  number  except  1  be  put  in  the  place  of  «,  and  y  be 
made  successively  1,  2,  3,  4,  a  suitable  value  may  be  found  for 
w,  which  shall  answer  the  conditions. 

If  a  be  made  1,  y  will  always  be  1,  whatever  value  be  given 
to  x;  for  all  powers,  as  well  as  all  roots  of  1,  are  1. 

But  if  any  number  greater  than  1  be  put  in  the  place  of  «,  y 
may  equal  any  number  whatever,  by  giving  x  a  suitable  value. 

Giving  a  value  to  a  then,  we  begin  and  make  y  successively 
1,  2,  3,  4,  &c.  and  these  numbers  will  form  the  first  column  or 
columns  of  powers  in  the  table.  Then  we  find  the  values  of 
X  corresponding  to  these  values  of  ?/,  and  write  them  in  the  se- 
cond column  against  the  values  of  y,  and  these  form  the  column 
of  exponents.  These  exponents  are  called  logarithms.  The 
first  column  is  usually  called  the  column  of  numbers,  and  the 
second,  the  column  of  logarithms.  The  number  put  in  the 
place  of  a,  is  called  the  base  of  the  table.  Whatever  number 
is  made  base  at  first,  must  be  continued  through  the  table. 

Observe  that  a*'  =  1  ;  therefore  whatever  base  be  used,  the 
logarithm  of  1  is  zero.  And  1  will  be  the  logarithm  of  the  base, 
for  a^  =:a. 

The  most  convenient  number  for  the  base,  and  the  one  ge- 
nerally used  in  the  tables,  is  10. 

10'  =:  1,  10'  =  10,  10'  =  100,  10'  =  1000,  10*  =  10000, 
10'  =z  100000,  10'  ==  1000000,  &c. 

Now  to  find  the  logarithm  of  2,  3,  4,  &c. 

Make  10^  =  2,  10'  =:  3,  10"^  =  4,  &c. 

For  all  numbers  between  1  and  10,  a?  must  be  a  fraction,  be- 
cause 10°  =  1  and  10'  =  10. 

Make  x  =z  — ,  then  it  becomes 


10'  =2. 


As  the  process  for  finding  the  value  ofz  in  this  equation  is 
long  and  rather  too  difficult  for  young  learners,  we  will  suppose 
it  already  found. 


244  Algebra,  XLIX. 

—  =  .30103  very  nearly. 

z         . 

Hence  10^»»»»»  =  2 very  nearly. 

To  understand  this,  we  must  suppose  10  raised  to  the  30103d 
power,  and  then  the  100000th  root  of  it  taken,  and  this  will 
differ  very  little  from  2.  The  number  .30103  is  the  logarithm 
of  2.  The  fractional  part  of  logarithms  is  always  expressed  in 
decimals. 

Having  the  logarithm  of  2,  we  may  find  the  logarithm  of  4 
by  doubling  it,  for  2^  =  4.  That  of  8  =  2'  is  found  by  tri- 
pling it,  and  so  on. 

The  logarithm  of  4  is  .30103  X  2=  .60206. 

The  logarithm  of  8  is  .30103  X  3  =  .90309. 

The  logarithm  of  16  is  .30103  X  4  =  1.20412,  &ic.     * 

Again  10^«»"  »""^  ==  .3  very  nearly. 

Hence  the  logarithm  of  3  is  .4771213. 

Since  2  X  3  =  6,  the  logarithm  of  6  is  found  by  adding  the 
logarithm  of  2  and  3  together. 

.30103  +  .4771213  =  .7781513  =  logarithm  of  6. 

Since  3^  =  9,  the  logarithm  of  9  is  found  by  multiplying  that 
of  3  by  2.  With  the  logarithms  of  2  and  3  the  logarithms  of  all 
the  powers  of  each,  and  of  all  the  multiples  of  the  two  may  be 
found. 

The  logarithm  of  5  may  be  found  by  subtracting  that  of  2 
from  that  of  10,  since  5  =:  y.     The  logarithm  of  10  is  1. 

1  —  .30103  =  .69897  =  log.  of  5. 
Now  all  the  logarithms  of  all  the  multiples  of  2,  3,  5,  and  10 
may  be  found.  Hence  it  appears  that  it  is  necessary  to  find 
the  logaritlmis  of  the  prime  numbers,  or  such  as  have  no  divi- 
sor except  unity,  by  trial  ;  and  then  the  logarithms  of  all  the 
compound  numbers  may  be  found  from  them. 

The  decimal  parts  of  the  logarithms  of  20,  30,  &c.  are  the 
same  as  those  of  2,  3,  4,  he.  For,  since  the  logarithm  of  10  is 
1  ;  that  of  100,  2  ;  that  of  1000,  3,  &;c.,  it  is  evident  that  add- 


XLIX.  Logarithms,  245 

ing  these  logarithms  to  the  logarithms  of  any  other  numbers, 
will  not  alter  the  decimal  part.  Hence  1  added  to  the  loga- 
rithm of  2  forms  that  of  20,  and  2  added  to  the  logarithm  of  2 
forms  that  of  200,  &c. 

Log.  2  =  .30103,  log.  20  =  1.30103,  log.  200  =z  2.30103, 
log.  2000^:3.30103. 

The  logarithm  of  25  is  1.39794  ;  that  of  250  =  25  x  10  is 
1  +  1.39794  =  2.39794 ;  that  of  2500  =  25  X  100  is  2  -f 
1.39794  =  3.39794. 

The  logarithms  of  all  numbers  below  10  are  fractions,  those 
of  all  the  numbers  between  10  and  100  are  1  and  a  fraction  ; 
those  of  all  numbers  between  100  and  1000  are  2  and  a  frac- 
tion ;  those  of  all  numbers  between  1000  and  10000  are  3  and  a 
fraction.  That  is,  the  whole  number  which  precedes  the  fraction 
in  the  logarithm  is  always  equal  to  the  number  of  figures  in  the 
number  less  one.  This  whole  number  is  called  the  index  or 
characteristic  of  the  logarithm.  Thus  in  the  logarithm  2.3576423, 
the  figure  2  is  the  characteristic  showing  that  it  is  the  loga- 
rithm of  a  number  consisting  of  three  figures  or  between  100 
and  1000. 

As  the  characteristic  may  always  be  known  by  the  number, 
and  the  number  of  figures  in  a  number  may  be  known  by  the 
characteristic,  it  is  usual  to  omit  the  characteristic  in  the  table, 
to  save  the  room.  It  is  useful  to  omit  it  too,  because  the  same 
fractional  part,  with  different  characteristics,  forms  the  loga- 
rithms of  several  different  numbers. 

The  logarithm  of  37  is  1.568202. 

Q«r  -J  rv'- 568202 

«>*    3  7   rr -__    1Q.568202. 

ro~~  '  ~~io      ~" 

The  logarithm  of  3.7  is  .568202,  which  is  the  same  as  that 
of  37,  with  the  exception  of  the  index. 

^_irz  =  37.62  =  — =  io^-5"4i9 

100  10' 

'VyM  1  A3.575419 

*lllf  =  3.762  =  — 1=:  lo-"''*'^' 

1000  10' 

21* 


246  Algebra.  XLIX. 

That  is,  all  numbers  which  are  tenfold,  the  one  of  the  other, 
have  the  same  logarithm. 

376200  has  for  its  logarithm     5.575419. 


37620 
3762         ' 
376.2 
37.62 
3.762 


4.575419. 
3.575419. 
2.575419. 
1.575419. 
0.575419. 


When  a  number  consists  of  whole  numbers  and  decimal 
parts,  we  find  the  fractional  part  of  the  logarithm  in  the  same 
manner  as  if  all  the  figures  of  the  number  belonged  to  the 
whole  number,  but  we  give  it  the  index  corresponding  to  the 
whole  number  only. 

In  most  tables  of  logarithms  they  are  carried  as  far  as  seven 
decimal  places.  Some  however  are  only  carried  to  five  or  six. 
The  disposition  of  the  tables  is  something  different  in  different 
sets,  but  they  are  generally  accompanied  with  an  explana- 
tion. When  one  set  of  tables  is  well  understood,  all  others  will 
be  easily  learned.  The  logarithms  for  the  following  examples 
may  be  found  in  any  table  of  logarithms.  They  are  used  here 
as  far  as  six  places. 


Examples. 

1.  Multiply  43  by  25. 

Find  43  in  the  column  of  numbers,  and  against  it  in  the  co- 
lumn of  logarithms  you  will  find  1.633468,  and  against  25  you 
will  find  1.397940.  Add  these  two  logarithms  together  and 
their  sum  is  the  logarithm  of  the  product. 

log.  43       ...       .       1.633468 

25  .       .       .  1.397940 


"  1075       ....       3.031408 

Find  this  logarithm  in  the  column  of  logarithms,  and  against 
it  in  the  column  of  numbers  you  find  1075  which  is  the  product 
of  43  multiplied  by  25.  The  index,  3,  shows  that  the  number 
must  consist  of  four  places. 


XLIX.  Logarithms.  247 

Let  the  learner  prove  the  results  at  first  by  actual  multipli- 
cation. 

2.  Multiply  2520  by  300. 

By  what  was  remarked  above,  the  logarithm  of  2520  is  the 
same  as  that  of  252  with  the  exception  of  the  index,  and  that 
of  300  is  the  same  as  that  of  3  except  the  index. 

Find  the  number  252  in  the  left  hand  column,  and  against  it 
in  the  second  column  you  find  .401401.  The  number  2520  con- 
sists of  four  places,  therefore  the  index  of  its  logarithm  must  be 
(4  —  1)  or  3.  The  logarithm  corresponding  to  300  is  .477121, 
and  its  index  must  be  2,  because  300  consists  of  three  places. 

log.  2520       ....       3.401401 

"  300  ...  2.477121 


*'  756000       ....       5.878522 

Find  this  logarithm,  and  against  it  in  the  column  of  numbers 
you  will  find  756;  but  the  index  5  shows  that  the  number 
must  consist  of  6  places  ;  therefore  three  zeros  must  be  annex- 
ed to  the  right,  which  makes  the  number  756000,  which  is  the 
product  of2520  by  300. 

3.  Multiply  2756  by  20. 

*  To  find  the  logarithm  of  2756,  find  in  the  column  of  num- 
bers 275,  and  at  the  top  of  the  table  look  for  6.  In  the  co- 
lumn under  6  and  opposite  275  you  find  .440279  for  the  deci- 
mal part  of  the  logarithm  of  2756.  The  characteristic  will 
be  3. 

log.  2756         .         .  .         3.440279 

"    20'       .         .         .  1.301030 


«    55120         .         .        .        4.741309 

Looking  in  the  table  for  this  logarithm,  against  551  you  will 
find  .741152  and  against  552  you  will  find  .741939.  The  lo- 
garithm .741309  is  between  these  two.  Against  551,  look 
along  in  the  other  columns.  In  the  column  under  2  you  find 
the  logarithm  required.     The  figures  of  the  number,  then,  are 

*  In  some  tables  the  whole  number  2756  may  be  found  in  the  left  hand  co- 
lumn. 


248  Algebra,  XLIX. 

5512,  but  the  characteristic  being  4,  the  number  must  consist 
of  five  places ;  hence  annexing  a  zero,  you  have  55120  for  the 
product  of  2756  by  20. 

4.  Divide  756342  by  27867. 

Both  these  numbers  exceed  the  numbers  in  the  tables,  still 
we  shall  be  able  to  find  them  w^ith  great  accuracy.  First  find 
the  logarithm  of  756300,  which  is  5.878694.  The  difference 
between  this  logarithm  and  that  of  756400  is  58.  The  differ- 
ence between  756300  and  756400  is  100,  and  the  difference 
between  756300  and  756342  is  42.  Therefore,  if  yVo  =  -42 
of  58  be  added  to  the  logarithm  of  756300,  it  will  give  the  lo- 
garithm of  756342  sufficiently  exact,  58  X  -42  =  24,  reject- 
ing the  decimals.  5.878694  +  24  =  5.878718.  The  58, 
and  consequently  the  24,  are  decim«ils  of  the  order  of  the  two 
last  places  of  the  logarithm,  but  this  circumstance  need  not  be 
regarded  in  taking  these  parts.  It  is  sufficient  to  add  them  to 
their  proper  place. 

The  table  generally  furnishes  means  of  taking  out  this  loga- 
rithm more  easily.  As  the  differences  do  not  often  vary  an 
unit  for  considerable  distance  among  the  higher  numbers,  the 
diflference  is  divided  into  ten  equal  parts,  (that  is,  as  equal  as 
possible,  the  nearest  number  being  used,  rejecting  the  decimal 
parts)  and  one  part  is  set  against  1,  two  parts  against  2,  &c.  in 
a  column  at  the  right  of  the  table. 

In  the  present  case,  then,  for  the  4  (for  which  we  are  to  take 
yV  of  58,)  we  look  at  these  parts  and  against  it  we  find  23,  and 
for  the  2  (for  which  we  must  take  y|^  of  58,)  we  find  11.  But 
11  is  y\,  consequently  to  obtain  y|^  we  must  take  y\  of  11 
which  is  1,  omitting  the  decimal.  The  operation  may  stand 
thus : 


log.  756300  5.878694 

3 
1 


yoOfdiff.  23 

_2_ 
1  0  0 


log.  756342  5.878718 

To  find  the  logarithm  of  27867,  proceed  in  the  same  manner, 
first  finding  that  of  27860,  and  then  adding  y'V  of  the  difference, 
which  will  be  found  at  the  right  hand,  as  above. 


L. 


log.  27860 
diff. 


Logarithms, 

4.444981 
109 


249 


log.  27867  4.445090 

From  log.  dividend         5.878718 
Subt.  log.  of  divisor       4.445090 

log.  of  quotient  27.141  1.43362S 

We  find  the  decimal  part  of  this  logaritnm  is  between 
.433610  and  .433770,  the  former  of  which  belongs  to  the  num- 
ber 2714,  and  the  latter  to  2715.  Subtract  433610  from 
433628,  the  remainder  is  18.  Looking  in  the  column  of  parts, 
the  number  next  below  18  is  17,  which  stands  against  1  or  yV 
of  the  whole  difference. 

Put  this  1  at  the  right  of  2714,  which  makes  27141.  The 
characteristic  1  shows  that  the  number  is  between  10  and  100. 
Therefore  the  quotient  is  27.141.  This  quotient  is  correct  to 
three  decimal  places. 

If  the  table  has  no  column  of  differences,  take  the  whole 
difference  between  .433610  and  .433770,  which  is  160  for  a 
divisor,  the  18  for  a  dividend,  annexing  one  or  more  zeros. 
One  place  must  be  given  to  the  quotient  for  each  zero. 


180 

160 

160 

.1 

5.  What  is  the  3d  power  of  25.7  ? 

log.  25.7 
Multiply  by  3 

Jim. 

1.409933 
3 

log.                           16974.6 

4.229799 
16974.6—. 

6.  What  is  the  3d  root  of  15  ? 
log.  15 

Ans. 

1.176091  ( 

log.  2.46621 

.392030 
2.4662+. 

L,  Since  a  fraction  consists  of  two  numbers,  one  for  the  nu- 
merator and  the  other  for  the  denominator,  the  logarithm  of  a 


250  Jllgehra.  L. 

fraction  must  consist  of  two  logarithms  ;  and  as  a  fraction  ex- 
presses the  division  of  the  numerator  by  the  denominator,  to 
express  this  operation  on  the  logarithms,  that  of  the  denomina- 
tor miist  be  represented  as  to  be  subtracted  from  the  nume- 
rator. 

The  logarithm  of  |  is  expressed  thus  : 

log.  3  —  log.  5  rr:  0.477121  —  0.698970. 
The  logarithm  of  a  fraction  whose  numerator  is  1,  may  be 
expressed  by  a  single  logarithm.     For  —  is  the  same  as  a~~. 


a^ 


If  we  would  express  the  logarithm  of  J  for  example  , 
1Q.477121  _  3^  consequently -1  ^^.^^^  =  Iq--'^'^''  =  J. 

That  is,  the  logarithm  of  J  is  the  same  as  the  logarithm  of  3, 
except  the  sign,  which  for  the  fraction  is  negative.  Any  frac- 
tion may  be  reduced  to  the  form  __,  but  the  denominator  will 

a" 

consist  of  decimals  or  still  contain  a  fraction. 

5 


If       1.666  + 

If  the  subtraction  be  actually  performed,  on  the  expression  of 
this  fraction  given  above,  it  will  be  reduced  to  the  logarithm 
of  a  fraction  of  this  form. 

0.477f21  —  0.698970  —  —  0.221849. 

The  number  corresponding  to  the  logarithm  0.221849  is  1.666 
-f-,  but  the  sign  being  negative,  shows  that  the  number  is 

1 


1.666  -f- 

The  logarithms  of  all  common  fractions  may  be  obtained  in 
either  of  the  above  forms,  but  they  are  extremely  inconvenient 
in  practice.  The  first  on  account  of  its  consisting  of  two  loga- 
rithms would  be  useless  as  well  as  inconvenient ;  because 
though  we  might  find  a  logarithm  corresponding  to  any  frac- 
tion, yet  in  performing  operations,  a  logarithm  would  never  be 
found  in  that  form  when  it  was  required  to  find  its  number. 


L.  Logarithms.  261 

The  second  form  is  inconvenient  because  it  is  negative,  and 
also  because  in  seeking  the  number  corresponding  to  the  loga- 
rithm, a  fraction  would  frequently  be  found  with  decimals  in 
the  denominator.  It  would  be  much  better  that  the  whole 
fraction  should  be  expressed  in  decimals.  If  the  fraction  is 
used  in  the  decimal  form,  the  logarithms  may  be  used  for  them 
almost  as  easily  as  for  whole  numbers. 

Suppose  it  is  required  to  find  the  logarithm  of  .5  or  y\. 

log.  5  -—  log.  10  =  0.698970  —  1 .  =  —  I  +  .698970. 

Suppose  it  is  required  to  find  the  logarithm  of  .05  or  yf  „. 

•I  n.69S970  —24-698970. 

log.  5  —  log.  100  —  0.698970—2  =  —  2  4-.698970. 

The  logarithms  of  10,  100,  1000,  &c.  always  being  whole 
numbers,  we  have  the  two  parts  distinct.  The  logarithm  of  .5 
is  the  same  as  that  of  5  except  that  it  has  the  number  1  joined 
to  it  with  the  sign  — ,  which  is  sufficient  to  distinguish  it,  and 
show  it  to  be  a  fraction.  The  logarithm  of  .05  also  is  the  same, 
except  that  —  2  is  joined  to  it.  That  is,  the  logarithm  of  the 
numerator  is  positive,  and  that  of  the  denominator  negative. 

This  negative  number  joined  to  the  positive  fractional  part, 
serves  as  a  characteristic,  and  is  a  continuation  of  the  principle 
shown  above ;  thus 

The  log.  500     is  2.698970 

"      "  50  1.698970 

"       «  5  0.698970 

«      «  .5  r.698970 

"      "  .05  2.698970 

The  logarithm  of  a  decimal  is  the  same  as  that  of  a  whole 
number  expressed  by  the  same  figures,  with  the  exception  of 
the  characteristic,  which  is  negative  for  the  fraction ;  being  — 
1  when  the  first  figure  on  the  left  is  tenths,  —  2  when  the  first 
is  hundredths,  &c.  It  is  convenient  to  write  the  sign  over  the 
characteristic  thus,  X,  ~2,  &c.  It  is  not  necessary  to  put  the 
sign  -\-  before  the  fractional  part,  for  this  will  always  be  un- 
derstood to  be  positive. 


^52  Algebra.  L. 

In  operating  upon  these  numbers,  the  same  rules  must  be  ob- 
served as  in  other  cases  where  numbers  are  found  connected 
with  the  signs  -\-  and  — . 

When  the  first  figure  of  the  fraction  is  tenths,  the  character- 
istic is  T,  when  the  first  is  hundredths,  the  characteristic  is  '% 

The  log.  of  .25  is  log.  25  —  log.  100 

=  1 .397940  —  2  —  —  2  -I-  1 .397940  =:T.397940. 

This  is  the  same  as  the  logarithm  of  25,  except  that  the  cha- 
racteristic T  shows  that  its  first  figure  on  the  left  is  lOths,  or 
one  place  to  the  right  of  units. 

Multiply  325  by.23. 

log.  325         .         .         .         2.511S83 
log.   .23  ...     ~r361728 

log.  74.75  ^?w.  .  .  1.873611 

Multiply  872  by  .097. 

log.  872         .         .  *         2.940516 

log.  .097  »  .  '2:986772 

log.  84.584  ^ws.         .         .  1.927288 

In  adding  the  logarithms,  there  is  1  to  carry  from  the  deci- 
mal to  the  units.  This  one  is  positive,  because  the  decimal 
part  is  so. 

Multiply  .857  by  .0093 

log.  .857         .         .  .         T.932981 

log.  .0093  .         .         .      "3".968483 


log.  .0079701  Ans.  .         .             "^901464 

Divide  75  by  .025. 

log.  75           .  .           .          1.875061 

log.  .025  .           .              "2:397940 


log.  dOOO  Ans.         .         .         .  2.477121 


L.  Logarithms,  253 

In  subtracting,  the 
gebraic  quantities. 

Divide  275  by  .047. 


In  subtracting,  the  negative  quantity  is  to  be  added,  as  in 
algebraic  quantities. 


log.  275 
log.  .047 

2.439333 
1:672098 

log.  5851.07  Am. 

3.767235 

•ivide  .076  by  830. 

log.  .076   "2:880814  = 
log.  830 

IT-f  1.880814 
2.919078 

log.  .0000915662  Ans.  .  ~5:961736 

In  order  to  be  able  to  take  the  second  from  the  first,  I  change 
the  characteristic  "2*  into  ^  +  1  which  has  the  same  value. 
This  enables  me  to  take  9  from  18,  that  is,  it  furnishes  a  ten  to 
borrow  for  the  last  subtraction  of  the  positive  part.  In  sub- 
tracting, the  characteristic  2  of  the  second  logarithm  becomes 
negative  and  of  course  must  be  added  to  the  other  negative. 

Divide   .735  by  .038. 

log.  .735  .  .  3;866287 

log.  .038         .  .  2.579784 


log.  19.3422  A*.  .  1.286503 

What  is  the  3d  power  of  .25  ^ 
log.  .25  .  .  T.397940 


log.  0.015625  Am,  T  +  1.193820  :=  "2:193820. 

What  is  the  third  root  of  0.015625  ? 

The  logarithm  of  this  number  is  "^193820.  This  character- 
istic ~2  cannot  be  divided  by  3,  neither  can  it  be  joined  with 
the  first  decimal  figure  in  the  logarithm,  because  of  the  differ- 
ent sign.  But  if  we  observe  the  operation  above  in  finding  the 
power,  we  shall  see,  that  in  multiplying  the  decimal  part  there 
was  1  to  carry,  which  was  positive,  and  after  the  multiplication 
was  completed,  the  characteristic  stood  thus,  T"  -f-  1  which  was 
22 


254  Algebra.  L. 

a'fterwards  rediiced  to  ^  Now  if  we  add  "i"  -|~  1  to  the  ^  in 
the  present  instance,  it  will  become  T  +  1,  and  at  the  sairie 
time  its  value  will  not  be  altered.  The  negative  part  of  the 
characteristic  will  then  be  divisible  by  3,  and  the  1  being  posi- 
tive may  be  joined  to  the  fractional  part. 

log.  .015625  "2.193820  =  T  -|-  1.193820(3 

log.  .25  Ans.  "r397940     . 

In  all  cases  of  extracting  roots  of  fractions,  if  the  negative  cha- 
racteristic is  not  divisible  by  the  number  expressing  tJie  root,  it  must 
be  made  so  in  a  similar  manner. 

If  the  characteristic  were  T  and  it  were  required  to  find  the 
fifth  root,  we  must  add  '2'  +  2  and  it  will  become  "5"  +  2. 

What  is  the  4th  root  of  .357  .? 

log.  .357  "r552668  r=:  T-h  3.552668(4 


log.  .77294  Ans.  ~r8S8147 

Any  common  fraction  may  be  changed  to  a  decimal  by  its 
logarithms,  so  that  when  the  logarithm  of  a  common  fraction  is 
required,  it  is  not  necessary  to  change  the  fraction  to  a  decimal 
previous  to  taking  it. 

It  is  required  to  find  the  logarithm  of  J  corresponding  to  } 
expressed  in  decimals. 

The  logarithm  of  2  being  0.30103,  that  of  i  will  be  —0.30103. 

Now  —  0.30103  -  —  1  +  1  —  .30103 

—  _- 1  -j.  (1  —  .30103)  =="169897. 

The  decimal  part  .69897  is  the  log.  of  5,  and  —  1  is  the  log. 
of  10  as  a  denominator.  Therefore  T69897  is  the  log.  of 
.5  =  i. 

Reduce  f  to  a  decimal. 

log.  5  0.69897  =  —  1  -f-  1.698970 

log.  8  0.903090 

log.  0.625  =  f  Ans.  T.795880 


X,,  Logarithnu,  255 

When  there  are  several  multiplications  and  divisions  to  be 
performed  together,  it  is  rather  more  convenient  to  perform  the 
whole  by  multiplication,  that  is,  by  adding  the  logarithms. 
This  may  be  effected  on  the  following  principle.  To  divide 
by  2  is  the  same  as  to  multiply  by  J  or  .5.  Dividing  by  5  is 
the  same  as  multiplying  by  ^  or  .2,  &ic. 

Suppose  then  it  is  required  to  divide  435  by  15.  Instead  of 
dividing  by  15  let  us  propose  to  multiply  by  j\.  First  find  the 
logarithm  of  j^^  reduced  to  a  decimal. 

log.  1  is  0  =  —  2  -f  2.000000 

log.  15  subtract  1.176091 


^^S'  T  s  ^^  forn™  of  a  decimal  2.823909 

log.  435  add  2.638489 


log.  29  =r  quotient  of  435  by  15  1.462398 

The  log.  of  yV  viz.  ^823909  is  called  the  Arithmetic  Compk- 
rnent  of  the  log.  of  1 5. 

The  arithmetic  complement  is  found  by  svhtracting  the  logarithm 
of  the  number  from  the  logarithm  of  I,  which  is  zero,  but  which  may 
always  be  represented  by  ~T  -(-  1,  "a"  -|-  2,  fyc.  It  must  always  be 
represented  by  such  a  number  that  the  logarithm  of  the  number  may 
be  subtracted  from  the  positive  part.  That  is,  it  must  always  be 
equal  to  the  characteristic  of  the  logarithm  to  be  subtracted,  plus  1 ; 
for  1  must  always  be  borrowed  from  it,  from  which  to  subtract  the. 
fractional  part. 

IX  is  required  to  find  the  value  of  x  in  the  following  equa- 
tion. 


(35  X  28  X  56.78\  ' 
387  X  2.896      / 


256  Algebra.  LI. 

1.544068 
1.447158 
1.754195 
a412289 
T538201 


log.  35 

log.  28 

log.  56.78 

log.  387 

2.587711     Arith.  Com. 

loff.  2.896 

0.461799       " 

1.695911 
3 

5.087733(5 


log.  10.4123  very  nearly  answer  1.017546 

I  multiply  by  3  to  find  the  3d  power,  and  divide  by  5  to  ob- 
tain the  5th  root. 

LI.  There  is  an  expedient  generally  adopted  to  avoid  the 
negative  characteristics  in  the  logarithms  of  decimals.  I  shall 
explain  it  and  leave  the  learner  to  use  the  method  he  likes  the 
best. 

1.  Multiply  253  by  .37. 

log.  .37  T.568202 

log.  253  2.403121 


log.  93.61  nearly  answer  1.971323 

Instead  of  using  the  logarithm  1.568202  in  its  present  form, 
add  10  to  its  characteristic  and  it  becomes  9.568202. 

log. .37  9.568202 

log.  253  2.403121 

11.971323 
Subtract  10. 


log.  93.61  as  above.  1.971323 

In  this  case  10  w^as  added  to  one  of  the  numbers  and  after- 
wards subtracted  from  the  result ;  of  course  the  answer  must 
be  the  same. 

2.  Multiply  .023  by  .976. 


LI.  Logarithms.  257 

Take  out  the  logarithms  of  these  numbers  and  add   10  to 
each  characteristic. 

log.  .023  8.361728 

log,  .976  9.989450 

18.351178 
Subtract  20 


log.  .0224473  nearly  ans.  2.351178 

We  may  observe  that,  in  this  way,  when  the  first  left  hand 
figure  is  tenths,  the  characteristic,  instead  of  being  T  is  9,  and 
when  the  first  figure  is  hundredths,  the  characteristic  is  8,  &c. 
That  is,  the  place  of  the  first  figure  of  the  number  reckoned 
from  the  decimal  point  corresponds  to  what  the  characteristic 
falls  short  of  10.  Whenever  in  adding,  the  characteristic  ex- 
ceeds 10,  the  ten  or  tens  may  be  omitted  and  the  unit  figure 
only  retained. 

In  the  first  example,  one  number  only  was  a  fraction,  viz. 
.37.  In  adding,  the  characteristic  became  11,  and  omitting 
the  10  it  became  1,  which  shows  that  the  product  is  a  number 
exceeding  10. 

In  the  second  example  both  numbers  were  fractions,  of  course 
each  characteristic  was  10  too  large.  In  adding,  the  charac- 
teristic became  18.  Now  instead  of  subtracting  both  tens  or 
20,  it  is  sufficient  to  subtract  one  of  them,  and  the  characteris- 
tic 8,  which  is  2  less  than  10,  shows  as  well  as  2~  would  do, 
that  the  product  is  a  fraction,  and  that  its  first  figure  must  be 
in  the  second  place  of  fractions  or  hundredth's  place. 

If  three  fractions  were  to  be  multiplied  together,  there  would 
be  three  tens  too  much  used,  and  the  characteristic  would  be 
between  20  and  30  ;  but  rejecting  two  of  the  tens,  or  20,  the 
remaining  figure  would  show  the  product  to  be  a  fraction,  and 
show  the  place  of  its  first  figure. 

3.  What  is  the  3d  power  of  .378  ? 

log.  .378        ....         9.577492 
Multiply  by  3 


28.732476 
log.  .05401  nearly  ans.  6.732476 

22  * 


258  Algebra,  LI. 

Multiplying  by  3  is  the  same  as  adding  the  number  twice  t<* 
itself.  The  characteristic  becomes  28,  but  omitting  two  of  the 
tens  or  20,  it  becomes  8,  which  shows  it  to  be  the  logarithm  of 
a  fraction  whose  first  place  is  hundredths. 

If  it  is  required  to  find  the  3d  root  of  a  fraction,  it  is  easy  to 
see,  that  having  taken  out  the  logarithm  of  the  fraction,  it  will 
be  necessary  to  add  two  tens  to  the  characteristic,  for  it  is  then 
considered  the  third  power  of  some  other  fraction,  and  in  rais- 
ing the  fraction  to  that  power,  two  tens  would  be  subtracted. 

In  the  last  example  the  logarithm  of  the  power  is  8.732476, 
but  in  order  to  take  its  3d  root,  it  will  be  necessary  to  add  the 
two  tens  which  were  omitted. 

For  the  second  root  one  ten  must  be  previously  added,  and 
for  the  fourth  root,  three  tens,  &c. 

4.  What  is  the  3d  root  of  .027  f 

log.  .027         .  .         V       :.         8.431364 

or  considered  as  a  3d  power  28.431364  (3 


log.  .3.  Ans,  .  .  .  .       9.477121 

5.  What  is  the  2d  root  of  .0016  ^ 

log.  .0016  .  .  7.204120 

or  considered  a  second  power  17.204120  (2 


log.  .04.  Ans.  8.602060 

In  dividing  a  whole  number  by  a  fraction,  if  10  be  added  to 
the  characteristic  of  the  dividend,  it  cancels  the  10  supposed 
to  be  added  to  the  divisor.  If  both  are  fractions  the  ten  in  the 
one  cancels  it  in  the  other ;  and  if  the  dividend  only  is  a  frac- 
tion, the  answer  will  of  course  be  a  less  fraction.  Consequent- 
ly in  division  the  results  will  require  no  alteration. 

6.  Divide  57  by  .018. 

log.  57  .  .  1.755875 

log.  .018  .  .         8.255272 

log.  3166.7  nearly  ans.  .  3.500603 


LI. 


Logarithms. 


^59 


Here  in  subtracting  I  suppose  10  to  be  added  to  the  first 
characteristic,  and  say  8  from  1 1 ,  &c. 


7.  Divide  .2172  by  .006. 

log.  .2172 
log.  .006 

log.  36.2  Ans. 


9.336860 
7.778151 

1.558709 


In  taking  the  arithmetical  complement,  the  logarithm  of  the 
lyimber  may  be  subtracted  immediately  from  10.  The  loga- 
rithm of  2  being  .301030,  its  arithmetical  complement  is 
T698970.  Adding  10  it  becomes  9.698970.  It  would  be  the 
same  if  subtracted  immediately  from  10  thus  10  —  .301030 
=  9.698970. 

8.  It  is  required  to  find  the  value  of  x  in  the  following  ex- 
pression : 


^  ___    17  /13.73  X  .0706\^ 
112  \  [253  / 


log.  13.73 

log.      .0706 

log.      .253     9.403121 

Sum 


Arith.  Com. 


Product  by  3 

Quotient  by  2 
log.  17 
log.  112     2.049218     Arith.  Com. 

log.  X  =     1.13835  nearly 


Find  the  value  of  a?  in  the  following  equations, 

X  =  /38.47  X  .463\^ 
\  .037  X  576  / 


1.137670 

8.848805 
0.596879 

0.583354 

r> 
O 

1.750062  (2 

0.875031 
1.230449 
7.950782 

0.056262 


260  Jilg^ra.  LII. 

3 


10. 


cc=z  1*5    /872  X  -0065^ 
417*  \.038  X4685/ 

x  =  —     /  873  \  '  /  278  \^ 
*  "^  476*   \  956  /  *  \1973/ 

12.  38^  =  683. 

Observe  that  the  2d  power  of  38  is  found  by  multiplying  the 
logarithm  of  38  by  2,  the  3d  power  by  multiplying  it  by  3,  &c. 
which  will  give  the  logarithm  of  the  result.  Hence  we  have 
the  following  equation;  the  logarithm  of  38  being  1.579784 
and  that  of  583  being  2.765669. 

X  X   1.579784  =  2.765669 

2.765669        ,  ^.^^^   , 

X  = =  1.75066  + 

1.579784 

The  value  of  x  is  found  by  dividing  one  logarithm  by  the 
other  in  the  same  manner  as  other  numbers.  It  might  be  done 
by  logarithms  if  the  tables  were  suiSciently  extensive  to  take 
out  the  numbers.  By  a  table  with  six  places  an  answer  cor- 
rect to  four  decimal  places  may  be  obtained. 

In  taking  out  the  logarithms  the  right  hand  figure  may  be 
omitted  without  affecting  the  result  in  the  first  four  decimals. 

log.  2.76567  .  .  0.441794 

log.  1.57978         .  .  .       0.198588 


log.  X  =  1.75064  +  .  0.243206 

13.  What  is  the  value  of  a?  in  the  equation  1537*  =  62  ? 
This  gives  first  1537  =  52^ 
This  may  now  be  solved  like  the  last. 

LII.     Questions  relating  to  Compound  Interest, 

It  is  required  to  find  what  any  given  principal  p  will  amount 
to  in  a  number  n  of  years,  at  a  given  rate  per  cent,  r,  at  com- 
pound interest. 

Suppose  first,  that  the  principal  is  $1,  or  £1,  or  one  unit  of 
money  of  any  kind. 


LII.  Compound  Interest.  261 

1    V  f 

The  interest  of  1  for  one  year  is  — -l—,  or  simply  r,  if  r  is 

^  100  ^^ 

considered  a  decimal.     The  amount  of  1  for  one  year  then, 
will  be  1  -{-  r.     The  amount  ofp  dollars  will  he  p  {I  -^  r). 

For  the  second  year,  p  (1  -\-  r)  will  be  the  principal,  and  the 
amount  of  1  being  (i  -\-  r),  the  amount  of  ^  (1  -j-  r)  will  be  p 
(I +r)  (I -\-r)  or p  {I -{-ry. 

For  the  third  year  p  (l  -\-  rf  being  the  principal,  the  amount 
will  be  p  (1  +  ry  (1  -{-  r)  or  p  (1  -f  r)^ 

For  n  years  then,  the  amount  will  he  p(l  -j-  r)". 

Putting  A  for  the  amount,  we  have 

A=p(l  +  r)«. 

This  equation  contains  four  quantities,  A,  p,  r,  and  n,  any 
three  of  which  being  given,  the  other  may  be  found. 

Logarithms  will  save  much  labour  in  calculations  of  this 
kind. 

Examples. 

1.  What  will  $753.37  amount  to  in  5|  years,  at  6  per  cent, 
compound  interest  f 

Herep  =  753.37,  r  =  .06,  and  w  =  5f. 

log.  l-fr  =  1,06  .  .  0.025306 

n=z  5| 


0.0031 63i 
3 

0.009490 
0.126530 


log.  (l+r)^^  ....  0.136020 

log,  753.37  *         .         :         .  2.877008 

log.  $1030.457  dns.  ,  3.013028 


262  Sgehra.  fifl. 

2.  What  principal  put  at  interest  will  amount  to  $5000  in 
13  years  at  5  per  cent,  compound  interest  ? 

By  the  above  formula 

A 


(i+ry 


log.  1  -f  r  =  1.05 


n  =z 

0.021189 
13 

.063567 
21189 

Subtract 
From 

.275457 
3.698970 

3.423513 

log.  A  =:  5000 

log.  p  :=  $2651.60  nearly  Atis. 

3.  At  what  rate  per  cent,  must  $378.57  be  put  at  compound 
interest,  that  it  may  amount  to  $500  in  5  years .'' 

Solving  the  equation  A  =^p  {\  -\-  ry  making  r  the  unknown 
quantity,  it  becomes 


r-f-l  =  f  — 


(f)' 


log.  v2  =  500 
log.  p  =.  378.57 

Dividing  by  n  =  5 


2.698970 
2.578146 

0.120824  (5 

0.024165 


log.  {r-\-\)  =  1.05722 
Consequently  r  —  0.05722  Ans. 

4.  In  what  time  will  $284.37  amount  to  750  at  7  per  cent.? 

Making  n  the  unknown  quantity,  the  equation  4  =i>  (1  +  ^)'* 
becom^es 

log.  2^  =  n  X  log.  (1  -f  r),  and 
P 


LII.  Compound  Interest,  263 

log.(l-f-r) 

log.  ^  =  750  .  .  .  .  2.875061 

log.  p  =  284.37  .  .  .  2.453881    ' 

log.  J:^ 0.421180 

P 
log.  1  +  r  =  1.07,  is  0.029384 

log.  0.421180 9.624467 

log.  0.029384  ....  8.468111 


log.  n  =  14.334  nearly  Ans.  1.156356 

5.  What  will  be  the  compound  interest  of  $947  for  4  years 
and  3  months  at  5J  per  cent.  ? 

6.  What  will  $157.63  amount  to  in  17  years  at  4f  per  cent.:" 

7.  A  note  was  given  the  15th  of  March  1804,  for  $58.46,  at 
the  rate  of  6  per  cent,  compound  interest ;  and  it  was  paid  the 
1 9th  of  Oct.  1823.     To  how  much  had  it  amounted  ? 

8.  A  note  was  given  the  13th  of  Nov.  1807,  for  $456.33,  and 
was  paid  the  23d  of  Sept.  1819.  The  sum  paid  was  $894.40. 
What  per  cejit.  was  allowed  at  compound  interest  ? 

9.  In  what  time  will  the  principal  p  be  doubled,  or  become 
2^,  at  6  per  cent,  compound  interest  ?  In  what  time  will  it 
be  tripled  ? 

JYote,  In  order  to  solve  the  above  question,  put  2  p  in  the 
place  of  A  for  the  first,  3  p  for  the  second,  and  find  the  value 
of  w. 

The  principles  of  compound  interest  will  apply  to  the  follow- 
ing questions  concerning  the  increase  of  population. 

10.  The  number  of  the  inhabitants  of  the  United  States  in 
A.  D.  1790  was  3,929,000,  and  in  1800,  5,306,000.  What  rate 
per  cent,  for  the  whole  time  was  the  increase  ?  What  per 
cent,  per  year  ? 


264  Algebra.  LIIL 

11.  Suppose  the  rate  of  increase  to  remain  the  same  for  the 
next  10  years,  what  would  be  the  number  of  inhabitants  in 
1810? 

12.  At  the  same  rate,  in  what  time  would  the  number  of  m- 
habitants  be  doubled  after  1 800  .'' 

13.  The  number  of  inhabitants  in  1810  by  the  census  was 
7,240,000.     What  was  the  annual  rate  of  increase  f 

14.  At  the  above  rate,  what  would  be  the  number  in  1820  ? 

1 5.  At  the  above  rate,  in  what  time  would  the  number  in 
1810  be  doubled .? 

16.  The  number  of  inhabitants  by  the  census  of  1820,  was 
9,638,000.  What  was  the  annual  rate  of  increase  from  1810 
tol820f 

17.  At  the  same  rate,  what  is  the  number  in  1825  ^ 

18.  At  the  same  rate,  what  will  be  the  number  in  1830? 

19.  At  the  same  rate,  in  what  time  will  the  number  in  1820 
be  doubled  ? 

20.  In  what  time  will  the  number  in  1820  be  tripled  ? 

21.  When  will  the  number  of  inhabitants,  by  the  rate  of  the 
last  census,  be  50,000,000  ? 


LIII.  1.  Suppose  a  man  puts  $10  a  year  into  the  savings 
bank  for  15  years,  and  that  the  rate  of  interest  which  the  bank 
is  able  to  divide  annually  is  5  per  cent.  How  much  money 
will  he  have  in  the  bank  at  the  end  of  the  15th  year  ? 

Suppose  a  =  the  sum  put  in  annually, 
r  =  the  rate  of  interest, 
t  =  the  time, 
A  =z  the  amount. 

According  to  the  above  rule  of  compound  interest,  the  sum 
a  at  first  deposited  will  amount  to  a  (r  -f-  1)S*  that  deposited 
the  second  year  will  amount  to  a  {r  -\-  1)*~' ;  that  deposited 


LlII.  Compound  Interest.  2^5 

the  third  year  will  amount  to  «  (r  -f-  1 V""* ;  that  deposited  the 
last  year  will  amount  to  a  {r  -|-  1)^     Hence  we  have 

.^  =  a  (r  -j-  ly  +  a  (r  -f  l)*"^  +  «  (r  -f  ly-^...  a  {r+  1)  .... 
=  al{r  +  iy  +  {r+ir^+{r+lY-\..,{r-{-l)] 

But  (r  -(-  1)S  (r  +  1)*~S  &'C.  is  a  geometrical  progressi<)n, 
whose  largest  term  is  (r  -j-  1)S  the  smallest  r  -\-  1,  and  the 
ratio  r  -f-  !•     The  sum  of  this  progression,  Art.  XL VII.  is 

(r  +  l)  [(r +  !)'-!] 

Hence  ^  =  ^+  1)  [(l+ 1)' -  0 
r 

The  same  result  may  be  obtained  by  another  course  of  rea- 
soning. 

The  amount  of  the  sum  a  for  one  year  is  a  -{-  a  r.  Adding 
a  to  this,  it  becomes  2  a  -{-  a  r.  "^ 

The  amount  of  this  at  the  end  of  another  year  is  2  a  -{-  ar 
-{-  2  a  r  -{-  a  r^,  or  2  a  -|-  3  a  r  +  «  r'.  Adding  a  to  this  it 
becomes 

S  a  -\-  S  a  r  -{-  a  ?'\ 

The  amount  of  this  for  1  year  is 

3  a  -f-  3  a  r  -f-  a  r*^  -f  3  a  r  -f-  3  a  /'"  -f-  a  r*, 

=z3a-\-6ar-\-4ar*-}'a  ?•', 

i=a(3-f6r  +  4r'-f  r'). 

This  is  the  amount  at  the  end  of  the  third  year  before  the 
addition  is  made  to  the  capital.  The  law  is  now  sufficiently 
manifest.  With  a  little  alteration,  the  quantity  3  -|-  6  r  -{-  4  r"* 
-j-  r^  may  be  rendered  the  4th  power  of  1  -}-  f-  The  three  last 
coefficients  are  already  right.  If  we  add  1  to  the  quantity  it 
becomes 

4  4-6  r  -[-  4  r'  -\-  r\ 

Multiply  this  by  r  and  it  becomes 

4  r  4-  6  r'  -f  4  r'  -^  r\ 
23 


266  Algehra,  LIII. 

Add  I  again  and  it  becomes 

1  -I-  4  r  +  6  r*  -f  4  r'  -f  r*. 

This  is  now  the  4th  power  of  1  -{-  r,  and  it  may  be  written 

(l+r^. 

Subtract  the  1  which  was  added  last,  and  it  becomes 

(l+r)'-l. 

Divide  this  by  r,  because  it  was  multiplied  by  r,  and  it  be- 
comes 

(i  +  O'-i 

r 

Subtract  1  again,  because  1  was  added  previous  to  multiply- 
ing by  r ;  and  it  becomes 

(l+'-r-l...   1  =  (l+ry-(l+r)  _  (l+r)  [(l+r)»-l] 
r  r  r 

Substitute  t  in  place  of  the  exponent  3,  and  multiply  by  a, 
and  it  becomes 

a(l+r)  [(l+r>-l]  ^^ 
r 
which  is  the  same  as  before. 

The  particular  question  given  above  may  now  be  solved  by 
logarithms,  using  this  formula. 

log.  (1  +  r)  =  1.05  .  .  .  0.021189 

Multiply  by,  ^  =15  .  .  15 


105945 
.21189 


log.  (1  +  r  y  =  2.079  .  .  .317835 

Subtract  1  1 


log.                        1.079  ...  0.033021 

log.  (1  +  r)            .         .  .          .  0.021189 

log.  a  =  10                  .  .          .         .  1.000000 

Arith.  Com.  log.  r  =  .05  .         .  1.301030 


Ans,  $226.59  ....         2.355240 


'>;■■ 


LIV.  Annuities.  26? 

2.  A  man  deposited  annually  $50  in  a  bank  from  the  time 
his  son  was  born,  until  he  was  20  years  of  age ;  and  it  was 
taken  out,  together  with  compound  interest  on  each  deposit  at 
3  per  cent.,  when  his  son  was  21  years  of  age,  and  given  to  him. 
How  much  did  the  son  receive  ? 

3.  How  much  did  the  bankers  gain  by  receiving  the  money, 
supposing  they  were  able  to  employ  it  all  the  time  at  6  per 
cent,  compound  interest  f 

4.  A  man  has  a  son  7  years  old,  and  he  wishes  to  give  him 
.$2000  when  he  is  21  years  old  ;  how  much  must  he  deposit 
annually  at  4  per  cent,  compound  interest,  to  be  able  to  do  it  ? 

5.  If  a  man  deposits  in  a  bank  annually  $35,  in  how  long  a 
time  will  it  amount  to  $500  at  6  per  cent,  compound  interest  ? 

6.  The  first  slaves  were  brought  into  the  American  Colonies 
in  the  year  1685.  Suppose  the  first  number  to  have  been  50, 
and  that  50  had  been  brought  each  year  for  100  years,  and  the 
rate  of  increase  3  per  cent.  How  many  would  there  have  been 
in  the  country  at  the  end  of  the  hundred  years  f 

LIV.     Annuities. 

1 .  A  man  died  leaving  a  legacy  to  a  friend  in  the  following 
manner  ;  a  sum  of  money  was  to  be  put  at  interest,  such  that, 
the  person  drawing  10  dollars  a  year,  at  the  end  of  15  years 
the  principal  and  interest  should  both  be  exhausted.  What 
sum  must  be  put  at  interest  at  6  per  cent,  to  fulfil  the  above 
condition  ? 

Let  the  learner  generalize  this  example  and  form  a  rule  ;  and 
then  solve  the  following  examples  by  it. 

,  2.  A  man  wishes  to  purchase  an  annuity  which  shall  afford 
him  $300  a  year  so  long  as  he  shall  live.  It  is  considered 
probable  that  he  will  live  30  years.  What  sum  must  he  de- 
posit in  the  annuity  office  t6  produce  this  sum,  supposing  he 
can  be  allowed  3  per  cent,  interest  .^ 

N.  B.  The  principal  and  interest  must  be  exhausted  at  the 
end  of  30  years 


26S  Algebra.  LIV. 

3.  If  the  man  mentioned  in  the  last  example  should  die  at 
the  end  of  18  years,  how  much  would  the  annuity  conipany 
gain  f 

4.  If  he  were  to  live  43  years,  how  much  would  the  company 


5.  A  man  purchases  an  annuity  for  life,  on  the  supposition 
that  he  shall  live  45  years,  for  $15000,  and  is  allowed  4  per 
cent,  interest.  How  much  must  he  draw  annually  that  the 
whole  may  be  exhausted  ? 

6.  A  man  has  property  to  the  amount  of  $35000,  which 
yields  him  an  income  of  5  per  cent.  His  annual  expenses  are 
$5000.     How  long  will  his  property  last  him  ? 

7.  The  number  of  slaves  in  the  United  States  in  1810  was 
1,191,000,  and  in  1820  the  number  was  1,531,000.  What  is 
the  number  at  present,  1 825,  allowing  the  rate  of  increase  to 
be  the  same  ? 

8.  There  is  a  society  established  in  the  United  States  for  the 
purpose  of  colonizing  the  free  people  of  colour.  Suppose  the 
slaves  to  be  emancipated  as  fast  as  this  society  can  transport 
them  away  ;  how  many  must  be  sent  away  annually,  that  the 
number  may  be  neither  increased  nor  diminished  ? 

9.  How  many  must  be  sent  away  annually  that  the  country 
may  be  cleared  in  100  years? 

10.  If  the  colonization  is  not  commenced  till  the  year  1840, 
supposing  the  rate  of  increase  to  remain  the  same  as  from  1810 
to  1820,  how  many  must  then  be  sent  away  annually,  that  the 
number  remaining  may  continue  the  same  ^ 

1 1 .  How  many  must  then  be  sent  away  annually,  that  the 
country  may  be  cleared  of  them  in  100  years  ? 

Miscellaneous  Examples, 

1 .  An  express  set  out  to  travel  240  miles  in  4  days,  but  in 
consequence  of  the  badness  of  the  roads  he  found  that  he  must 
go  5  miles  the  second  day,  9  the  third,  and  14  the  fourth,  less 
than  the  first.    How  many  miles  must  he  travel  each  day  f 


Miscellaneous  Examples.  269 

2.  Two  workmen  received  the  same  sum  for  their  labour ; 
but  if  one  had  received  27  shillings  more  and  the  other  19 
shillings  less,  then  one  w^ould  have  received  just  three  times  as 
much  as  the  other.     What  did  they  receive  ? 

3.  Two  persons,  A  and  B  worked  together,  A  worked  15 
and  B  18  days,  and  they  received  equal  sums  for  their  work. 
But  if  A  had  worked  11  i  and  B  14  days,  then  A  would  have 
received  35  shillings  more  than  B.  What  was  the  daily  wages 
of  each  ? 

4.  Two  merchants  entered  into  a  speculation,  by  which  one 
gained  54  dolkrs  more  than  the  other.  The  whole  gain  was 
49  dollars  less  than  three  times  the  gain  of  the  less  What 
were  the  gains  ^ 

%  5.  A  man  bought  a  piece  of  cloth  for  a  certain  sum,  and  on 
measuring  it,  found  that  it  cost  him  8  dollars,  but  if  there  had 
been  4  yards  more,  it  would  have  cost  him  only  $7  per  yard. 
How  many  yards  were  there  ? 

6.  Divide  the  number  46  into  two  such  parts,  that  one  of 
them  being  divided  by  7,  and  the  other  by  3,  the  quotients  may 
together  be  equal  to  10. 

7.  A  farm  of  864  acres  is  divided  between  3  persons.  C  has 
as  many  acres  as  A  and  B  together  ;  and  the  portions  of  A 
and  B  are  in  the  proportion  of  5  to  11.  How  many  acres  had 
each  f 

8.  There  are  two  numbers  in  the  proportion  of  ^  to  f ,  the 
first  of  which  being  increased  by  4  and  the  second  by  6,  they 
will  be  in  the  proportion  of  f  to  J.     What  are  the  numbers  ^ 

9.  A  farmer  has  a  stack  of  hay,  from  which  he  sells  a  quan- 
tity, which  is  to  the  quantity  remaining  in  the  proportion  of  4 
to  5.  He  then  uses  15  loads,  and  finds  that  he  has  a  quantity 
left,  which  is  to  the  quantity  sold  as  1  to  2.  How  many  loads 
did  the  stack  at  first  contain  f 

10.  There  are  3  pieces  of  cloth,  whose  lengths  are  in  the 
proportion  of  3,  5,  and  7  ;  and  8  yards  being  cut  oif  from  each, 
the  whole  quantity  is  diminished  in  the  proportion  of  15  to  11. 
What  was  the  length  of  each  piece  at  first  ? 

11.  The  number  of  days  that  4  workmen  were  employed 
mere  severally  as  the  numbers  4,  5,  6,7  5  their  wages  were  the 

23* 


270  Algebra. 

same,  viz.  3  shillings,  and  the  sum  received  by  the  first  and 
second  was,  36  shillings  less  than  that  received  by  the  third  and 
fourth.     How  much  did  each  receive  ? 

12.  There  are  two  numbers,  the  greater  of  which  is  three 
times  the  less  ;  and  the  sum  of  their  second  powers  is  five  times 
the  sum  of  the  numbers.     What  are  the  numbers  ^ 

13.  What  two  numbers  are  those,  of  which  the  less  is  to  the 
greater  as  2  to  3  ;  and  w^hose  product  is  six  times  the  sum  of 
the  numbers  ? 

14.  There  are  two  boys,  the  difference  of  whose  ages  is  to 
their  sum  as  2  to  3,  and  their  sum  is  to  their  product  as  3  to  5. 
What  are  their  ages  ? 

15.  A  detachment  of  soldiers  from  a  regiment  being  ordered 
to  march  on  a  particular  service,  each  company  furnished  4 
times  as  many  men  as  there  were  companies  in  the  regiment ; 
but  these  being  found  insufficient,  each  company  furnished 
three  more  men,  when  their  number  was  found  to  be  increased 
in  the  proportion  of  17  to  16.  How  many  companies  were 
there  in  the  regiment  ? 

16.  Find  two  numbers  which  are  in  the  proportion  of  8  to  5, 
and  whose  product  is  360. 

17.  A  draper  bought  2  pieces  of  cloth  for  $31.45,  one  being 
50  and  the  other  65  cents  per  yard.  He  sold  each  at  an  ad- 
vanced price  of  12  cents  per  yard,  and  gained  by  the  whole 
$6.36.     What  were  the  lengths  of  the  pieces  ? 

18.  Two  labourers,  A  and  B,  received  $43.85  for  their  wages ;^ 
A  having  been  employed  15,  and  B  14  days;  and  A  received 
for  working  four  days  $3.25  more  than  B  for  3  days.  What 
were  their  daily  wages  f 

10.  Having  bought  a  certain  quantity  of  brandy  at  19  shil- 
lings per  gallon,  and  a  quantity  of  rum  exceeding  that  of  the 
brandy  by  9  gallons,  at  15  shillings  per  gallon,  I  find  that  I 
paid  one  shilling  more  for  the  brandy  than  for  the  rum.  How 
many  gallons  were  there  of  each  ? 

20.  Two  persons,  A  and  B,  have  each  an  annual  income  of 
$1200.  A  spends  every  year  $120  more  than  B,  and  at  the 
end  of  4  years  the  amount  of  their  savings  is  equal  to  one  year's 
income  of  either.     What  does  each  spend  annually  ? 


Miscellaneous  Examples.  271 

.21.  In  a  naval  engagement,  the  number  of  ships  taken  was 
T'more,  and  the  number  burnt  was  2  fewer,  than  the  number 
sunk  ;  1 5  escaped,  and  the  fleet  consisted  of  8  times  the  num- 
ber sunk.     Of  how  many  did  the  fleet  consist  ? 

22.  A  cistern  is  filled  in  50  minutes  by  3  pipes,  one  of  which 
conveys  10  gallons  more,  and  the  other  8  gallons  less  than  the 
third,  per  minute.  The  cistern  holds  1820  gallons.  How 
much  flows  through  each  pipe  per  minute .'' 

23.  A  farm  of  750  acres  is  divided  between  three  persons, 
A,  B,  and  C.  C  has  as  much  as  A  and  B  both,  wanting  10 
acres ;  and  the  shares  of  A  and  B  are  to  each  other  in  the 
proportion  of  7  to  3.     How  many  acres  has  each  ? 

24.  A  certain  sum  of  money  being  put  at  interest  for  8 
months,  amounts  to  $772.50.  The  same  sum  put  out  at  the 
same  rate  for  15  months  amounts  to  792.1875,  Required 
the  sum  and  the  rate  per  cent. 

25.  From  two  casks  of  equal  size  are  drawn  quantities  which 
are  in  the  proportion  of  5  to  8  ;  and  it  appears  that  if  20  gal- 
lons less  had  been  drawn  from  the  one  which  now  contains  the 
less,  only  f  as  much  would  have  been  drawn  from  it  as  from 
the  other.     How  many  gallons  were  drawn  from  each  ? 

26.  There  are  two  pieces  of  land,  which  are  in  the  form  of 
rectangular  parallelograms.  The  longer  sides  of  the  two  are 
in  the  proportion  of  6  to  1 1 ,  and  the  adjacent  sides  of  the  less 
are  in  the  proportion  of  3  to  2.  The  whole  distance  round  the 
less  is  135  yards  greater  than  the  longer  side  of  the  larger 
piece.  Required  the  sides  of  the  less,  and  the  longer  side  of 
the  greater. 

27.  A  person  distributes  forty  shillings  amongst  fifty  people, 
giving  some  9d.  and  the  rest  I5d.  each.  How  many  were 
there  of  each  f 

28.  Divide  the  number  49  into  two  such  parts,  that  the  quo- 
tient of  the  greater  divided  by  the  less,  may  be  to  the  quotient 
of  the  less  divided  by  the  greater  as  f  to  f . 

29.  A  person  put  a  certain  sum  to  interest  for  5  years,  at  6 
per  cent,  simple  interest,  and  found  that  if  he  had  put  out  the 
same  sum  for  8  years  at  4 J  per  cent,  he  would  have  received 
^60  more.     What  was  the  sum  put  out  ? 


272  Algebra. 

30.  A  regiment  of  militia  containing  830  men  is  to  be  raised 
from  three  towns,  A,  B,  and  C  The  contingents  of  A  and  B 
are  in  the  proportion  of  3  to  5  ;  and  of  B  and  C  in  the  propor- 
tion of  6  to  7.     Required  the  numbers  raised  by  each. 

31.  At  what  time  between  6  and  7  o'clock  are  the  hour  and 
minute  hands  of  a  watch  together  ? 

32.  There  is  a  number  consisting  of  two  digits,  the  second 
of  which  is  greater  than  the  first ;  and  if  the  number  be  divided 
by  the  sum  of  the  digits,  the  quotient  will  be  4  ;  but  if  the  digits 
be  inverted  and  that  number  divided  by  a  number  greater  by  2 
than  the  difference  of  the  digits,  the  quotient  will  be  14.  Re- 
quired the  number. 

33.  There  is  a  fraction  whose  numerator  being  tripled,  and 
the  denominator  diminished  by  3,  the  value  becomes  §  ;  but  if 
the  denominator  be  doubled  and  the  numerator  increased  by 
2,  its  value  becomes  4.     Required  the  fraction. 

34.  A  merchant  bought  a  hogshead  of  wine  for  f  100.  A 
few  gallons  having  leaked  out,  he  sold  the  remainder  for  the 
original  sum,  thus  gaining  a  sum  per  cent,  on  the  cost  of  it, 
equal  to  twice  the  number  of  gallons  which  leaked  out.  How 
many  gallons  did  he  lose  ? 

35.  There  are  two  pieces  of  cloth,  differing  in  length  4 
yards ;  the  first  is  worth  as  many  shillings  per  yard  as  the  se- 
cond contains  yards  ;  the  second  is  worth  as  many  shillings  per 
yard  as  the  first  contains  yards ;  and  both  pieces  are  worth 
£72.  10s.     How  many  yards  does  each  contain  ^ 

36.  A  merchant  bought  a  piece  of  cloth  for  $180,  and  sell- 
ing it  at  an  advance  of  $1  a  yard  on  the  cost,  he  gained  15 
per  cent.     Required  the  number  of  yards. 

37.  There  are  two  rectangular  pieces  of  land,  whose  lengths 
are  to  each  other  as  3 : 2,  and  surfaces  as  5  : 3  ;  the  smaller  one 
is  20  rods  wide.     What  is  the  width  of  the  other  ? 

38.  There  is  a  cistern  to  be  filled  with  a  pump,  by  a  man 
and  a  boy  working  at  it  alternately  ;  the  man  would  do  it  in 
15  hours,  the  boy  in  20.  They  filled  it  in  16  hours  48  minutes. 
How  long  did  each  work  ^ 

39.  In  a  bag  of  money  there  is  a  certain  number  of  eagles, 
as  many  quarter  eagles,  f  the  number  of  half  eagles,  together 


Miscellaneous  Examples.  273 

with  dollars  sufficient  to  make  up  the  number  of  coins  equal  to 
^  of  the  value  of  the  whole  in  dollars ;  and  the  number  of  ea- 
gles and  dollars  diminished  by  2,  is  half  the  number  of  coins. 
What  is  the  number  of  coins  of  each  sort  ? 

40.  Suppose  a  man  owes  $1000,  what  sum  shall  he  pay 
daily  so  as  to  cancel  the  debt,  principal  and  interest,  at  the  end 
of  a  year,  reckoning  it  at  6  per  cent,  simple  interest  ? 

41.  A  merchant  bought  two  pieces  of  linen  cloth,  containing 
together  120  yards.  He  sold  each  piece  for  as  many  cents  per 
yard  as  it  contained  yards,  and  found  that  one  brought  him  in 
only  4  as  much  as  the  other.  How  many  yards  were  there  in 
each  piece  .'* 

42.  A  criminal  having  escaped  from  prison,  travelled  10 
hours  before  his  escape  was  known.  He  was  then  pursued  so 
as  to  be  gained  upon  3  miles  an  hour.  After  his  pursuers  had 
travelled  8  hours,  they  met  an  express  going  at  the  same  rate 
as  themselves,  who  met  the  criminal  2  hours  and  24  min.  be- 
fore. In  what  time  from  the  commencement  of  the  pursuit  will 
they  overtake  him  ? 

43.  A  and  B  enter  into  partnership  with  a  joint  stock  of 
$900.  A's  capital  was  employed  4  months,  and  B's  7  months. 
When  the  stock  and  gain  were  divided,  A  received  $512,  and 
B  $469.     What  was  each  man's  stock  ? 

44.  A  gentleman  bought  a  rectangular  lot  of  valuable  land, 
giving  10  dollars  for  every  foot  in  the  perimeter.  If  the  same 
quantity  had  been  in  a  square,  and  he  had  bought  it  in  the  same 
way,  it  would  have  cost  him  $33  less ;  and  if  he  had  bought  a 
square  piece  of  the  same  perimeter  he  would  have  had  12^  rods 
more.     What  were  the  dimensions  of  the  piece  he  bought  ? 

45.  A  and  B  put  to  interest  sums  amounting  together  to  800 
dollars.  A's  rate  of  interest  was  1  per  cent,  more  than  B's, 
his  yearly  interest  f  of  B's  ;  and  at  the  end  of  10  years  his  prin- 
cipal and  simple  interest  amounted  to  f  of  B's.  What  sum  was 
put  at  interest  by  each,  and  at  what  rate  ? 

46.  Two  messengers,  A  and  B,  were  despatched  at  the  same 
time  to  a  place  90  miles  distant ;  the  former  of  whom  riding  one 
mile  an  hour  more  than  the  other,  arrived  at  the  end  of  his  jour- 
ney an  hour  before  him.   At  what  rate  did  each  travel  per  hour  ? 

47.  A  and  B  lay  out  some  money  on  speculation.  A  dis- 
poses of  his  bargain  for  $11,  and  gains  as  much  per  cent,  as  B 


274  Algebra. 

lays  out ;  B's  gain  is  ^'36,  and  it  appears  that  A  gains  four 
times  as  much  per  cent,  as  B.     Required  the  capital  of  each  ? 

48.  A  and  B  hired  a  pasture,  into  which  A  put  four  horses, 
and  B  as  many  as  cost  him  1 8  shilhngs  a  week.  Afterwards  B 
put  in  two  additional  horses,  and  found  that  he  must  pay  20 
shillings  a  week.     At  what  rate  was  the  pasture  hired  ? 

49.  A  vintner  draws  a  certain  quantity  of  wine  out  of  a  full 
vessel  that  holds  256  gallons ;  and  then  filling  the  vessel  with 
water,  draws  off  the  same  quantity  of  liquor  as  before,  and  so 
on  for  four  draughts,  when  there  were  only  81  gallons  of  pure 
wine  left.     How  much  wine  did  he  draw  each  time  ? 

50.  Three  merchants,  A,  B,  and  C,  made  a  joint  stock,  by 
which  they  gained  a  sum  less  than  that  stock  by  $80.  A's 
share  of  the  gain  was  $60,  his  contribution  to  the  stock  was  $17 
more  than  B's.  B  atid  C  together  contributed  $325.  How 
much  did  each  contribute  ? 

51.  A  grocer  sold  80lb.  of  mace  and  lOOlb.  of  cloves  for 
£^b  ;  but  he  sold  60  pounds  more  of  cloves  for  £20  than  he 
did  of  mace  for  £10.     What  was  the  price  of  a  pound  of  each  ? 

52.  A  and  B,  165  miles  distant  from  each  other,  set  out  with 
a  design  to  meet.  A  travels  one  mile  the  first  day,  two  the 
second,  three  the  third,  and  so  on  ;  B  travels  20  miles  the  first 
day,  18  the  second,  16  the  third,  and  so  on.  In  how  many 
days  will  they  meet .'' 

53.  A  and  B  engage  to  reap  a  field  for  $20  ;  and  as  A  alone 
could  reap  it  in  9  days,  they  promise  to  complete  it  in  5  days. 
They  found  however  that  they  were  obliged  to  call  in  C  to  as- 
sist them  for  the  two  last  days,  in  consequence  of  which,  B  re- 
ceived f  of  a  dollar  less  than  he  otherwise  woiild  have  done. 
In  what  time  could  B  or  C  alone  reap  the  field  ? 

54.  A  mercer  bought  a  piece  of  silk  for  $54  ;  and  the  num- 
ber of  shillings  which  he  paid  for  a  yard  was  a  of  the  number 
of  yards.  How  many  yards  did  he  buy,  and  what  was  the 
price  of  a  yard  ^ 

55.  The  fore  wheel  of  a  carriage  makes  6  revolutions  more 
than  the  hind  wheel  in  going  120  yards ;  but  if  the  periphery 
of  each  wheel  be  increased  one  yard,  it  will  make  only  4  revo- 
lutions more  than  the  hind  wheel  in  the  same  space.  Requir- 
ed the  circumference  of  each. 


Miscellaneous  Examples.  275 

56.  There  are  three  numbers,  the  difference  of  whose  differ- 
ences is  5 ;  the  sum  of  the  numbers  is  20,  and  their  continued 
product  130.     Required  the  numbers. 

57.  From  two  towns,  C  and  D,  two  travellers,  A  and  B,  set 
out  to  meet  each  other ;  and  it  appeared  that  when  they  met, 
B  had  gone  35  miles  more  than  |  of  the  distance  that  A  had 
travelled  ;  but  from  their  rate  of  travelling,  A  expected  to 
reach  C  in  20|  hours ;  and  B  to  reach  D  in  30  hours.  Re- 
quired the  distance  from  C  to  D. 

58.  Two  men,  A  and  B,  entered  into  a  speculation,  to  which 
B  contributed  $15  more  than  A.  After  four  months,  C  was 
admitted,  who  added  $50  to  the  stock ;  and  at  the  end  of  12 
months  from  C's  admission  they  had  gained  $159;  when  A 
withdrawing  received  for  principal  and  gain  $88.  What  did 
he  originally  subscribe  ? 

59.  The  number  of  deaths  in  a  besieged  garrison  amounted 
to  6  daily  ;  and  allowing  for  this  diminution,  their  stock  of  pro- 
visions was  sufficient  to  last  8  days.  But  on  the  evening  of  the 
sixth  day,  100  men  were  killed  in  a  sally,  and  afterwards  the 
mortality  increased  to  10  daily.  Supposing  the  stock  of  pro- 
visions unconsumed  at  the  end  of  the  sixth  day  to  support  6 
men  for  61  days;  it  is  required  to  find  how  long  it  would  sup- 
port the  garrison,  and  the  number  of  men  alive  when  the  pro- 
visions were  exhausted. 

60.  There  was  a  cask  containing  20  gallons  of  brandy ;  a 
certain  quantity  of  this  was  drawn  off  into  another  cask  of  equal 
size,  and  this  last  filled  with  water,  and  afterwards  the  first 
cask  was  filled  with  the  mixture.  It  now  appears  that  if  6| 
gallons  of  the  mixture  be  drawn  off  from  the  first  into  the  se- 
cond cask,  there  will  be  equal  quantities  of  brandy  in  each. 
Required  the  quantity  of  brandy  first  drawn  off. 

61.  From  two  towns,  C  and  D,  which  were  at  the  distance 
of  396  miles,  two  persons,  A  and  B,  set  out  at  the  same  time, 
and  meet  with  each  other,  travelling  as  many  days  as  are  equal 
to  the  difference  of  the  number  of  miles  they  travelled  per  day ; 
when  it  appeared  that  A  has  travelled  216  miles.  How  many 
miles  did  each  travel  per  day  1 

62.  A  tailor  bought  a  piece  of  cloth  for  $200,  from  which 
he  cut  5  yards  for  his  own  use,  and  sold  the  remainder  for 


276  Algebra, 

$115,  gaining  75  cents  per  yard.      How  many  yards  wierc 
there,  and  what  did  it  cost  him  per  yard  ? 

G3.  There  is  a  rectangular  field  containing  10  acres,  1  quar- 
ter, 5  rods,  and  the  length  of  it  exceeds  the  breadth  by  12  rods. 
Required  the  dimensions  of  the  field. 

64.  A  man  travelled  96  miles,  and  then  found  that  if  he  had 
travelled  2  miles  faster  per  hour,  he  should  have  been  8  hours 
less  in  performing  the  same  journey.  At  what  rate  per  hour 
did  he  travel  ? 

65.  A  regiment  of  soldiers,  consisting  of  900  men,  is  formed 
into  two  squares,  one  of  which  has  6  men  more  in  a  side  than 
the  other.  What  is  the  number  of  men  in  a  side  of  each 
square  ? 

66.  A  and  B  travelled  on  the  same  road  and  at  the  same 
rate  from  Huntingdon  to  London.  At  the  50th  mile  stone 
from  London,  A  overtook  a  drove  of  geese  which  were  pro- 
ceeding at  the  rate  of  three  miles  in  two  hours ;  and  two 
hours  afterwards  met  a  stage  waggon,  which  was  moving  at 
the  rate  of  9  miles  in  4  hours.  B  overtook  the  same  drove  of 
geese  at  the  45th  mile  stone,  and  met  the  same  stage  waggon 
exactly  forty  minutes  before  he  came  to  the  31st  mile  stone. 
Where  was  B  when  A  reached  London  ? 

67.  Two  men,  A  and  B,  bought  a  farm  consisting  of  200 
acres,  for  which  they  paid  $200  each.  On  dividing  the  land, 
A  says  to  B,  if  you  will  let  me  have  my  part  in  the  situation 
which  I  shall  choose,  you  shall  have  so  much  more  land  than  I, 
that  mine  shall  cost  75  cents  per  acre  more  than  yours.  B  ac- 
cepted the  proposal.  How  much  land  did  each  have,  and 
what  was  the  price  of  each  per  acre  ^ 

68.  A  person  bought  two  cubical  stacks  of  hay  for  4 1£ ;  each 
of  them  cost  as  many  shillings  per  solid  yard  as  there  were  yards 
in  a  side  of  the  other,  and  the  greater  stood  on  more  ground  than 
the  less  by  9  square  yards.     What  was  the  price  of  each  ? 

69.  Two  partners,  A  and  B,  dividing  their  gain  $60  B 
took  $20 ;  A's  money  was  in  trade  4  months,  and  if  the  num- 
ber 50  be  divided  by  A's  money,  the  quotient  will  give  the 
number  of  months  that  B's  money,  which  was  $100,  continued 
in  trade.  What  was  A's  money,  and  how  long  did  B's  con- 
tinue in  trade  ^ 

END, 


v:; 


OK  . 


IMPROVED  SCHOOL,  BOOKS. 


Colburn's  First  Lessons,  or,  Intellectual  Arithmciic. 

THE  merits  of  this  little  work  are  so  well  known,  and  so  highly  appreciat- 
ed in  Boston  and  its  vicinity,  that  any  recommendation  of  it  is  unnecessary, 
except  to  those  parents  and  teachers  in  the  country,  to  whom  it  has  not  been 
introduced.  To  such  it  may  be  interesting  and  important  to  be  informed,  that 
the  system  of  which  this  work  gives  the  elementary  principles,  is  founded  on 
this  simpk  maxim  ;  that,  children  should  be  instructed  in  every  science,  just  so 
fast  as  they  can  understand  it.  In  conformity  with  this  principle,  the  book 
commences  with  examples  so  simple,  that  they  can  be  perfectly  comprehended 
and  performed  mentally  by  children  of  four  or  five  years  of  age  ;  having  per- 
formed these,  the  scholar  will  be  enabled  to  answer  the  more  difficult  questions 
which  follow.  He  will  find,  at  every  stage  of  his  progress,  that  what  he  has 
already  done  has  perfectly  prepared  him  for  what  is  at  present  required.  This 
will  encourage  him  to  proceed,  and  will  afford  him  a  satisfaction  in  his  study, 
which  can  never  be  enjoyed  while  performing  the  merely  mechanical  operation 
of  ciphering  according  to  artificial  rules. 

This  method  entirely  supersedes  the  necessity  of  any  rules,  and  the  book 
contains  none.  The  scholar  learns  to  reason  correctly  respecting  all  combina- 
tions of  numbers  ;  and  if  he  reasons  correctly,  he  must  obtain  the  desired  re- 
sult. The  scholar  who  can  be  made  to  understand  how  a  sum  should  be  done, 
needs  neither  book  nor  instructor  to  dictate  how  it  must  bo  done. 

This  admirable  elementary  Arithmetic  introduces  the  scholar  at  once  to  that 
simple,  practical  system,  which  accords  with  the  natural  operations  of  the  hu- 
man mind.  AH  that  is  learned  in* this  way  is  precisely  what  will  be  found  es- 
sential in  transacting  the  ordinary  business  of  life,  and  it  prepares  the  way,  in 
the  best  possible  manner,  for  the  more  abstruse  investigations  which  belong  to 
maturer  age.  Children  of  five  or  six  years  of  age  will  be  able  to  make  consi- 
derable progress  in  the  science  of  numbers  by  pursuing  this  simple  method  of 
studying  it,  and  it  will  uniformly  be  found  that  this  is  one  of  the  most  useful 
and  interesting  sciences  upon  which  their  minds  can  be  occupied.  By  using 
this  work  children  may  be  farther  advanced  at  the  age  of  nine  or  ten,  than  they 
can  be  at  the  age  of  fourteen  or  fifteen  by  the  common  method.  Those  who 
have  used  it,  and  are  regarded  as  competent  judges,  have  uniformly  decided 
that  more  can  be  learned  from  it  in  one  year,  than  can  be  acquired  in  two  years 
from  any  other  treatise  ever  published  in  America.  Those  who  regard  econo- 
my in  time  and  money,  cannot  fail  of  holding  a  w'ork  in  high  estimation  which 
will  afford  these  important  advantages. 

Colburn's  First  Lessbns  are  accompanied  with  such  instructions  as  to  the 
proper  mode  of  using  them,  as  will  relieve  parents  and  teachers  from  any  em- 
barrassment. The  sale  of  the  work  has  been  so  extensive  that  the  publishers 
have  been  enabled  so  to  reduce  its  price,  that  it  is,  at  once,  the  cheapest  and 
the  best  Arithmetic  in  the  country. 


Improved  School  Books. 

Colburn^s  Sequel. 

THIS  work  consists  of  two  parts,  in  the  first  of  which  the  author  has  given  a 
great  variety  of  questions,  arranged  according  to  the  method  pursued  in  the 
First  Lessons  ;  the  second  part  consists  of  a  few  questions,  with  the  solution  of 
them,  and  such  copious  illustrations  of  the  principles  involved  in  the  examples 
in  the  first  part  of  the  work,  that  the  whole  is  rendered  perfectly  intelligible. 
The  two  parts  are  designed  to  be  studied  together.  The  answers  to  the  ques- 
tions in  the  first  part  are  given  in  a  Key,  which  is  published  separately  for 
the  use  of  instructers.  If  the  scholar  find  any  sum  difiicull,  he  must  turn  to  the 
principles  and  illustrations,  given  in  the  second  part,  and  these  will  furnish  ali 
the  assistance  that  is  needed. 

The  design  of  this  arrangement  is  to  make  the  scholar  understand  his  subject 
thoroughly,  instead  of  performing  his  sums  by  rule. 

The  First  Lessons  contain  only  exaniples  of  numbers  so  small,  that  they  can 
be  solved  without  the  use  of  a  slate.  The  Sequel  commences  with  small  and 
simple  combinations,  and  proceeds  gradually  to  the  more  extensive  and  varied, 
and  the  scholar  will  rarely  have  occasion  for  a  principle  in  arithmetic  which  is 
not  fully  illustrated  in  this  work. 


Colbmii's  Introduction  to  Algebra. 

THOSE  who  are  competent  to  decide  on  the  merits  of  this  work  consider  it 
equal,  at  least,  to  either  of  the  others  composed  by  the  same  author. 

The  publishers  cannot  desire  that  it  should  have  a  higher  commendation. 
The  science  of  Algebra  is  so  much  simplified,  that  children  may  proceed  with 
ease  and  advantage  to  the  study  of  it,  as  soon  as  they  have  finished  the  preced- 
ing treatises  on  arithmetic.  The  same  method  is  pursued  in  this  as  in  the  au- 
thor's other  works;  every  thing  is  made  plain  as  he  proceeds  with  his  subject. 

The  uses  which  are  performed  by  this  science  give  it  a  high  claim  to  more 
general  attention.  Few  of  the  more  abstract  mathematical  investigations  can 
be  conducted  without  it ;  and  a  great  proportion  of  those,  for  which  arithmetic 
is  used,  would  be  performed  vvitli  much  greater  facility  and  accuracy  by  an  al- 
gebraic process. 

The  study  of  Algebra  is  singularly  adapted  to  discipline  the  mind,  and  give  it 
direct  and  simple  modes  of  reasoning,  and  it  is  universally  regarded  as  one  of 
the  most  pleasing  studies  in  which  the  mind  can  be  engaged. 


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